Definite Integrals - Class 12 Commerce Applied Mathematics - Extra Questions

$$\displaystyle \alpha = \int_0^{1} { \left (e^{ \left (9x + 3 \tan^{-1}x \right) } \right) } \left( \dfrac{12+9x^2}{1+x^2} \right) dx$$
Let $$z = 9x+3\tan^{-1} x\Rightarrow dz = \cfrac{12+9x^2}{1+x^2}dx$$
$$\therefore \displaystyle \alpha =  \int_0^{9+\frac{3\pi}{4}} e^z dz =e^{ 9+\frac{3\pi}{4}} -1$$
$$\displaystyle\Rightarrow \log(\alpha+1) =9+\frac{3\pi}{4}$$
$$\therefore \Rightarrow \log(\alpha+1) -\frac{3\pi}{4}=9$$

$$I=\int \dfrac{\cos \sqrt{x}}{\sqrt{x}}dx$$

$$\left(\int_{0}^{\pi/6} sec^{2} x dx\right)^2=(\dfrac1{\sqrt3})^2=\dfrac13=0.33$$

$$y={ 2 }^{ 2 }{ 3 }^{ 2x }{ 5 }^{ -5 }{ 7 }^{ -7 }\\ y={ 2 }^{ 2 }{ 5 }^{ -5 }{ 7 }^{ -7 }{ 3 }^{ 2x }\\ \dfrac { dy }{ dx } ={ 2 }^{ 2 }{ 5 }^{ -5 }{ 7 }^{ -7 }{ 3 }^{ 2x }\log { 2x } \quad (2)\\ \dfrac { dy }{ dx } ={ 2 }^{ 3 }{ 5 }^{ -5 }{ 7 }^{ -7 }{ 3 }^{ 2x }\log { 2x } \quad$$

$$\displaystyle \int{\sin x\sin(2x)\,dx} = \int{\sin x(2\sin x\cos x)\,dx}$$

$$\begin{array}{l}I = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x{{\cos }^3}xdx} \\I = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x{{\cos }^2}x\cos xdx} \\I = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x\left( {1 - {{\sin }^2}x} \right)\cos xdx} \\\left\{ \begin{array}{l}Let,\sin x = t\, ;\, \cos xdx = dt\\x = 0\, ;\, t = 0\\x = \frac{\pi }{2}\, ;\, t = 1\end{array} \right\}\\I = \int\limits_0^1 {{t^2}\left( {1 - {t^2}} \right)dt} \\I = \left[ {\cfrac{{{{\sin }^3}x}}{3} - \cfrac{{{{\sin }^5}x}}{5}} \right]_0^1 = \cfrac{1}{3} - \cfrac{1}{5} = \cfrac{2}{{15}}\end{array}$$

Let $$I = \int {\dfrac{{1 + x}}{{\sqrt {1 - {x^2}} }}dx}$$

$$\,\int\limits_0^\pi  {\dfrac{{n\sin 2x.\sin \left( {\pi /2\cos x} \right)}}{{2x - \pi }}dx}$$

$$\displaystyle \int\limits_0^{\pi /4} {{{\sin x\cos x} \over {{{\cos }^4}x + {{\sin }^4}x}}dx}$$

$$= \displaystyle \int\limits_0^{\pi /4} {{{\sin x\cos x} \over {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x}}dx}$$

$$= \displaystyle \int\limits_0^{\pi /4} {{{\sin x\cos x} \over {1 - {\pi  \over 2}2{{\sin }^2}x{{\cos }^2}x}}dx}$$

$$= \displaystyle \int\limits_0^{\pi /4} {{{\sin x\cos x} \over {1 - {{{{\sin }^2}2x} \over 2}}}}$$

$$= \displaystyle \int\limits_0^{\pi /4} {{{2\sin x\cos x} \over {2 - {{\sin }^2}x2x}}dx}$$

$$= \displaystyle \int\limits_0^{\pi /4} {{{\sin 2x} \over {1 + {{\cos }^2}x}}dx}$$

Let $$t = \cos 2x$$

$$dt =  - 2\sin 2xdt$$

$$=  - {1 \over 2}\displaystyle \int\limits_1^0 {{{dt} \over {1 + {t^2}}} =  - {1 \over 2}\left[ {{{\tan }^{ - 1}}t} \right]_1^0}$$

$$=  - {1 \over 2}\left[ {{{\tan }^{ - 1}}0 - {{\tan }^{ - 1}}1} \right] =  - {1 \over 2}\left[ {0 - {\pi  \over 4}} \right]$$

$$= {\pi  \over 8}$$

Let $$5x+6=t$$ , then on differentiating the equation we get, $$5dx=dt$$ .

Now,

$$\displaystyle I=\int\limits_{0}^{2\pi}\cos^5x dx$$       ......(1)

$$\displaystyle\int\limits_{0}^{1} \dfrac{x}{x^2+1}dx$$

Consider $$\displaystyle I=\int\limits_{0}^{\pi/2} \cos^2x \ dx$$           .........(1)

Consider the given integral.

$$I=\displaystyle \int_{1}^{3}{\dfrac{\cos \left( \log x \right)}{x}}dx$$

 

Let $$t=\log x$$

$$dt=\dfrac{dx}{x}$$

 

Therefore,

$$I=\displaystyle \int_{0}^{\log 3}{\cos tdt}$$

$$I=\left[ \sin t \right]_{0}^{\log 3}$$

$$I=\left[ \sin \left( \log 3 \right)-\sin 0 \right]$$

$$I=\sin \left( \log 3 \right)$$

 

Hence, this is the answer.

$$\int_0^\infty  {\dfrac{{\log x}}{{1 + {x^2}}}dx}$$

$$I=\int\limits_{0}^{1} \sqrt{x(1-x)}dx$$

$$\displaystyle \int_0^{\pi/4} \tan^3 x\sec x dx$$

$$=\displaystyle \int_0^{\pi/4} \tan^2 x\tan x\sec x dx$$

$$=\displaystyle \int_0^{\pi/4} (\sec^2 x-1)\tan x\sec x dx$$

Let $$\sec x = t$$ , then $$\tan x\sec x dx = t$$
When $$x=0, t=1;x=\pi/4,t=\sqrt{2}$$

$$=\displaystyle \int_{1}^{\sqrt{2}}(t^2-1)dt$$

$$=\left[\dfrac{t^3}{3}-t\right]_1^\sqrt{2}$$

$$=\dfrac{2\sqrt{2}-1}{3}-(\sqrt{2}-1)$$

$$=\boxed{\dfrac{2-\sqrt{2}}{3}}$$

$$\int ^{2/3}_0 \cfrac{1}{4+9x^2}dx$$

We have,

$$\int_0^{2}x\sqrt{x+2}dx$$

$$\displaystyle \int_{-\pi/2}^{\pi/2}\dfrac{1-\cos2x}{2} dx$$

$$=\left[\dfrac{x}{2}-\dfrac{\sin2x}{4}\right]_{-\pi/2}^{\pi/2}$$

$$=\boxed{\dfrac{\pi}{2}}$$

Now,

Let $$I=\displaystyle \int_{0}^{\pi /2}  {1+\sin x} dx$$  

$$\displaystyle \int_{0}^{\pi /2} {1+\cos x} dx$$  

$$\displaystyle \int_0^1 x^2+2x dx\\\displaystyle \int _0^1 x^2dx +\int _0^1 2x dx \\\left.\dfrac {x^3}{3}+x^2\right|_0^1\|$$

$$\displaystyle \int_{0}^{3}x^2 \ dx$$  

$$\displaystyle\int_{0}^{1}1-x^{2}dx$$

$$\displaystyle\int_{0}^{\pi/2} \sin x \cos x\ dx$$  

$$\displaystyle \int_0^{1}(3x^2+2x)dx$$

$$\displaystyle \int_{0}^{2}(x^2+2x +1)dx$$

$$I=\displaystyle \int_{0}^{5} x^2dx$$

Let $$\displaystyle I=\int _{ 0 }^{ 1 } \frac { 1 }{ \left( 1+x^{ 2 } \right) ^{ 3/2 } } dx$$

Let $$\displaystyle I=\int _{ 0 }^{ \infty  } \frac { dx }{ \left( x+\sqrt { 1+x^{ 2 } }  \right) ^{ n } }$$
Substitute $$x=\tan  \theta \Rightarrow dx=\sec ^{ 2 }{ \theta  } d\theta$$
$$\displaystyle I=\int _{ 0 }^{ \pi /2 } \frac { \sec ^{ 2 } \theta d\theta  }{ \left( \tan  \theta +\sec  \theta  \right) ^{ n } } =\int _{ 0 }^{ \pi /2 } \frac { \cos ^{ n-2 } \theta d\theta  }{ \left( 1+\sin  \theta  \right) ^{ n } }$$
Using $$\int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( a+b-x \right) dx }$$
$$\displaystyle I=\int _{ 0 }^{ \pi /2 } \frac { \sin ^{ n-2 } \theta d\theta  }{ \left( 1+\cos  \theta  \right) ^{ n } } =\int _{ 0 }^{ \pi /2 } \frac { \left( 2\sin  \theta /2\cos  \theta /2 \right) ^{ n-2 } }{ \left( 2\cos ^{ 2 } \theta /2 \right) ^{ n } } d\theta$$
$$\displaystyle =\frac { 1 }{ 4 } \int _{ 0 }^{ \pi /2 } \frac { \sin ^{ n-2 } \theta /2 }{ \cos ^{ n+2 } \theta /2 } d\theta =\frac { 1 }{ 4 } \int _{ 0 }^{ \pi /2 } \frac { \sin ^{ n-2 } \theta /2 }{ \cos ^{ n-2 } \theta /2 } d\theta .\frac { 1 }{ \cos ^{ 4 } \theta /2 } d\theta$$
$$\displaystyle I=\frac { 1 }{ 4 } \int _{ 0 }^{ \pi /2 } \tan ^{ n-2 } \theta \left( 1+\tan ^{ 2 } \frac { \theta  }{ 2 }  \right) .\sec ^{ 2 } \frac { \theta  }{ 2 } .\sec ^{ 2 } \frac { \theta  }{ 2 } d\theta$$
Substitute $$\displaystyle \tan  \frac { \theta  }{ 2 } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 } \frac { \theta  }{ 2 } d\theta =dt$$
$$\displaystyle I=\frac { 1 }{ 4 } \int _{ 0 }^{ 1 } t^{ n-2 }\left( 1+t^{ 2 } \right) .2dt=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \left( t^{ n-2 }+t^{ n } \right) dt$$
$$\displaystyle =\frac { 1 }{ 2 } \left[ \frac { t^{ n-1 } }{ n-1 } +\frac { t^{ n+1 } }{ n+1 }  \right] ^{ 1 }_{ 0 }=\frac { 1 }{ 2 } \left[ \frac { 1 }{ n-1 } +\frac { 1 }{ n+1 }  \right] =\frac { n }{ n^{ 2 }-1 }$$
Hence $$b=35,a=6$$  
Therefore $$b-a=29$$

Let $$I=\int _{ 0 }^{ \pi  }{ x\log { sin { x }  }  }$$   ...(1)
using $$\int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ b }{ f\left( a+b-x \right) } dx$$
$$\therefore I=\int _{ 0 }^{ \pi  }{ \left( \pi -x \right) \log { \left( sin{ \left( \pi -x \right)  } \right)  }  } dx$$
$$=\int _{ 0 }^{ \pi  }{ \left( \pi -x \right) \log { sin{ x } }  } dx$$   ...(2)
From (1) and (2)
$$2I$$ $$\displaystyle=\pi \int _{ 0 }^{ \pi  }{ \log { sin{ x } }  } dx==\pi \left( \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { sin{ x } }  } dx+\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { \left( sin { \left( \pi -x \right)  }  \right)  }  } dx \right)$$
using $$\int _{ 0 }^{ 2a }{ f\left( x \right)  } dx=\int _{ 0 }^{ a }{ f\left( x \right) +f\left( 2aq-x \right)  } dx$$
$$\displaystyle\therefore 2I=2\pi \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { sin{ x } }  } dx$$
$$\displaystyle\Rightarrow I=\pi \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { sin{ x } }  } dx$$   ...(3)
Now using $$\int _{ 0 }^{ a }{ f\left( x \right)  } dx=\int _{ 0 }^{ a }{ f\left( a-x \right)  } dx$$
$$\displaystyle \therefore I=\pi \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { sin{ \left( \dfrac { \pi  }{ 2 } -x \right)  } }  } dx$$
$$=\pi \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { cos { x }  }  } dx$$   ...(4)
From (3) and (4)
$$\Rightarrow 2I=\pi \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ \log { \left( sin { x } cos { x }  \right)  }  } dx=\pi \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { sin { 2x }  }  } dx-\pi \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log {  { 2 }  }  } dx$$
Substitue $$2x=t\Rightarrow 2dx=dt$$
$$\displaystyle\therefore 2I=\dfrac { \pi  }{ 2 } \int _{ 0 }^{ \pi  }{ \log { { sin { t }  } }  } dt-{ \left[ \pi x\log { 2 }  \right]  }_{ 0 }^{ \dfrac { \pi  }{ 2 }  }$$
$$\displaystyle 2I=\dfrac { 2\pi  }{ 2 }\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { sint }  } dt-\dfrac { { \pi  }^{ 2 } }{ 2 } \log { 2 }$$  
$$\displaystyle\Rightarrow 2I=\pi \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { { sin{ x } } }  } dx-\dfrac { { \pi  }^{ 2 } }{ 2 } \log { 2 }$$
$$\displaystyle \therefore I=\dfrac { { \pi  }^{ 2 } }{ 2 } \log { \dfrac { 1 }{ 2 }  }$$
Hence, $$k+m+n=2+2+2=6$$

$$\displaystyle \int_{0}^{\pi /2}\frac{\cos x dx}{1-\sin ^{2}x+\sin ^{4}x}$$

Put $$\sin x=t$$
$$\cos xdx=dt$$
When $$x=0 \Rightarrow t=0$$  
When $$x=\dfrac{\pi}{2} \Rightarrow t=1$$

$$I=\displaystyle \int _{ 0 }^{ 1 } \frac { dt }{ { t }^{ 4 }-{ t }^{ 2 }+1 }$$

$$I=\displaystyle \frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { \frac { 2 }{ { t }^{ 2 } } dt }{ { t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } -1 }$$

$$I=\displaystyle \frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { \frac { 1 }{ { t }^{ 2 } } +1+\frac { 1 }{ { t }^{ 2 } } -1dt }{ { t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } +2-2-1 }$$

$$I=\displaystyle \frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { \frac { 1 }{ { t }^{ 2 } } +1dt }{ (t-\frac { 1 }{ { t } } )^{ 2 }+(1)^{ 2 } } -\frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { 1-\frac { 1 }{ { t }^{ 2 } } dt }{ (t+\frac { 1 }{ { t } } )^{ 2 }-(\sqrt { 3 } )^{ 2 } }$$

Put $$t-\dfrac { 1 }{ { t } } =u$$ in first integral
$$\Rightarrow (1+\dfrac { 1 }{ { t }^{ 2 } } )dt=du$$

Put $$t+\dfrac { 1 }{ { t } } =v$$ in second integral
$$\Rightarrow (1-\dfrac { 1 }{ { t }^{ 2 } } )dt=dv$$

$$I=\displaystyle \frac { 1 }{ 2 } \int _{ -\infty  }^{ 0 } \frac { du }{ u^{ 2 }+1^{ 2 } } +\frac { 1 }{ 2 } \int _{ 2 }^{ \infty  } \frac { dv }{ v^{ 2 }-(\sqrt { 3 } )^{ 2 } }$$

$$I=\displaystyle \frac { 1 }{ 2 } [\tan ^{ -1 }{ u } ]_{ -\infty  }^{ 0 }+\frac { 1 }{ 4\sqrt { 3 }  } [\log { |\frac { v-\sqrt { 3 }  }{ v+\sqrt { 3 }  } | } ]_{ 2 }^{ \infty  }$$

$$I=\displaystyle \frac { \pi  }{ 4 } +\frac { 1 }{ 4\sqrt { 3 }  } \log { |\frac { 2+\sqrt { 3 }  }{ 2-\sqrt { 3 }  } | }$$

$$\displaystyle \int_0^1\frac {x}{x^2+1}dx$$
Let $$x^2+1=t\Rightarrow 2x dx=dt$$
When $$x=0, t=1$$ and when $$x=1, t=2$$
$$\therefore\displaystyle  \int_0^1\frac {x}{x^2+1}dx=\frac {1}{2}\int_1^2\frac {dt}{t}$$
$$\displaystyle =\frac {1}{2}[\log|t|]_1^2$$
$$\displaystyle =\frac {1}{2}[\log 2-\log 1]=\frac {1}{2}\log 2$$

Let $$I = \displaystyle\int _{ 1 }^{ 5 }{ \frac { dx }{ \sqrt { 2x-1 }  }  }$$ .

$$\int _{ 0 }^{ \pi /4 }{ \dfrac { dx }{ \cos ^{ 3 }{ x } \sqrt { 2sin2x }  }  } =\int _{ 0 }^{ \pi /4 }{ \dfrac { dx }{ \cos ^{ 3 }{ x } \sqrt { 4sinx\cos { x }  }  }  } \\ =\int _{ 0 }^{ \pi /4 }{ \dfrac { dx }{ \cos ^{ 4 }{ x } \sqrt { 4\tan { x }  }  }  } =\dfrac { 1 }{ 2 } \int _{ 0 }^{ \pi /4 }{ \dfrac { \sec ^{ 4 }{ x }  }{ \sqrt { \tan { x }  }  } dx } \\ =\int _{ 0 }^{ \pi /4 }{ \dfrac { \sec ^{ 2 }{ x } (1+\tan ^{ 2 }{ x } ) }{ \sqrt { \tan { x }  }  } dx } \\ t=\tan { x } \Rightarrow dt=\sec ^{ 2 }{ x } dx\\ \Rightarrow \dfrac { 1 }{ 2 } \int _{ 0 }^{ \pi /4 }{ \dfrac { \sec ^{ 2 }{ x } (1+\tan ^{ 2 }{ x } ) }{ \sqrt { \tan { x }  }  } dx } =\dfrac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ ({ t }^{ -\frac { 1 }{ 2 }  }+{ t }^{ \frac { 3 }{ 2 }  }) } dt\\ ={ [{ t }^{ \frac { 1 }{ 2 }  }+\dfrac { { t }^{ \frac { 5 }{ 2 }  } }{ 5 } ] }_{ 0 }^{ 1 }=\dfrac { 6 }{ 5 }$$

The integral can be evaluated by transforming it into polar coordinates.

Apply u-substitution u= $${ x }^{ 2 }$$

Given : $$\int _{ -\pi  }^{ \pi  }{ \cfrac { \cos ^{ 2 }{ x } dx }{ (1+{ a }^{ x }) }  }$$

Given : $$\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ \sqrt { 1-\sin { 2x }  } dx }$$

$$\int_{1}^{2} [\alpha^{2} + (4 - 4\alpha) x + 4x^{3}]dx \leq 12$$

$$\int_{0}^{a}x\ dx \leq a + 4$$

$$=\cfrac { 1 }{ 9 } \displaystyle \int _{ 0 }^{ 2/3 }{ \cfrac { dx }{ \left( \cfrac { 4 }{ 9 } +{ x }^{ 2 } \right)  }  }$$

$$\displaystyle \int_{\frac{\pi}{2}}^{\pi}\frac{1-sinx}{1-cosx}dx=\int \dfrac{1}{1-\cos x}dx-\int \dfrac{\sin x}{1-\cos x}dx=\int \dfrac{1}{2\sin^2\dfrac{x}{2}}dx-\int \dfrac{\sin x}{1-\cos x}dx\\=\dfrac12 \int cosec^2\dfrac x2-\int \dfrac{\sin x}{1-\cos x}dx=\dfrac12 *2*\left[\cot \dfrac x2\right]^{\pi}_{\frac \pi2}-\int \dfrac{\sin x}{1-\cos x}dx\\=[\cot \dfrac{\pi}{2}-\cot\dfrac{\pi}{4}]-\int \dfrac{\sin x}{1-\cos x}dx=0-1-\int \dfrac{\sin x}{1-\cos x}dx$$

$$T=\overset{10n^2}{\underset{0}{\int}}\dfrac{3e^{3x}+2e^{2x}}{e^{3x}+e^{2x}+1}dx$$

$$\int_0^{\frac{\pi}{4}}(\cos 2x)^{\frac{3}{2}}\cos x dx$$

$$\begin{array}{l}\int_0^{\pi /4} {\frac{{dx}}{{{{\cos }^3}x\sqrt {2\sin 2x} }}}  = \frac{6}{5}\\ = \int_0^{\pi /4} {\frac{{se{c^3}xdx}}{{\sqrt {4\sin x\cos x} }}}  = \frac{6}{5}\\ = \int_0^{\pi /4} {\frac{{se{c^3}xdx}}{{2\cos x\sqrt {\tan x} }}}  = \frac{6}{5}\\ = \frac{1}{2}\int_0^{\pi /4} {\frac{{se{c^4}xdx}}{{\sqrt {\tan x} }}}  = \frac{6}{5}\\put,y = \tan x\\dy = {\sec ^2}xdx\\\frac{1}{2}\int\limits_0^1 {\frac{{\left( {1 + {y^2}} \right)}}{{\sqrt y }}dy} \\ = \frac{1}{2}\left( {\left[ {2\sqrt y } \right]_0^1 + \left[ {\frac{{{y^{5/2}}}}{{5/2}}} \right]_0^1} \right)\\ = \frac{1}{2}\left( {2 + \frac{2}{5}} \right) = \frac{6}{5}\end{array}$$

We have,

$$I=\int_{2}^{6}{\sqrt{\left( 6-x \right)\left( x-2 \right)}}\,dx$$

$$I=\int_{2}^{6}{\sqrt{6x-12-{{x}^{2}}+2x}}\,dx$$

$$I=\int_{2}^{6}{\sqrt{8x-12-{{x}^{2}}}}\,dx$$

$$I=\int_{2}^{6}{\sqrt{8x-12-16+16-{{x}^{2}}}}\,dx$$

$$I=\int_{2}^{6}{\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{\left( 4-x \right)}^{2}}}}\,dx$$

 

Let

$$t=4-x$$

$$-dt=dx$$

 

Therefore,

$$I=-\int_{2}^{-2}{\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{t}^{2}}}}\,dt$$

 

We know that

$$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\dfrac{1}{2}x\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$$

 

Therefore,

$$I=-\left[ \dfrac{1}{2}t\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{t}^{2}}}+\dfrac{1}{2}{{\left( 2\sqrt{7} \right)}^{2}}{{\sin }^{-1}}\left( \dfrac{t}{2\sqrt{7}} \right) \right]_{2}^{-2}$$

$$=-\left( \dfrac{1}{2}\left( -2 \right)\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{\left( -2 \right)}^{2}}}+\dfrac{1}{2}{{\left( 2\sqrt{7} \right)}^{2}}{{\sin }^{-1}}\left( \dfrac{-2}{2\sqrt{7}} \right) \right)+$$

          $$\left( \dfrac{1}{2}\left( 2 \right)\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{\left( 2 \right)}^{2}}}+\dfrac{1}{2}{{\left( 2\sqrt{7} \right)}^{2}}{{\sin }^{-1}}\left( \dfrac{2}{2\sqrt{7}} \right) \right)$$

$$I=- \left( -\sqrt{28-4}+\dfrac{1}{2}\times 28{{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{7}} \right) \right)+\left( \sqrt{28-4}+\dfrac{1}{2}\times 28{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{7}} \right) \right)$$

$$I=-\left( -\sqrt{24}+14{{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{7}} \right) \right)+\left( \sqrt{24}+14{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{7}} \right) \right)$$

$$I=2\sqrt{24}-14{{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{7}} \right)+14{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{7}} \right)$$

 

Hence, the value is $$2\sqrt{24}-14{{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{7}} \right)+14{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{7}} \right)$$ .

Let $$x = tan\theta$$

$$\int _{ 0 }^{ 2 }{ x\sqrt { x+2 }  } dx$$

Putting $$\cos x = t$$  

$$\begin{array}{l} \int _{ 2 }^{ 4 }{ \frac { { dx } }{ x }  }  \\ =\left[ { \log  x } \right] _{ 2 }^{ 4 } \\ =\left( { \log  4-\log  2 } \right)  \\ =\left( { 2\log  2-\log  2 } \right)  \\ =\log  2. \end{array}$$

$$\displaystyle I = \int_z^y (6x^3 + 5x)dx$$
$$\displaystyle =6\int_z^y x^3dx + 5\int_z^y x \, dx$$
$$=6\left[\dfrac{x^4}{4}\right]^y_z + 5\left[\dfrac{x^2}{2}\right]^y_z$$
$$=\dfrac{3}{2} \left[y^4 - z^4\right] + \dfrac{3}{2} [y^2 - z^2] = \dfrac{3}{2}[240] + \dfrac{5}{2}[12]$$
$$=360+30 = 390$$

$$\displaystyle \int_0^{\pi/4} x \tan x  \sec^2x dx$$

$$3 = \ e^{ \ln3}$$
$$\implies 3^x = e^{\ x\ln3}$$

$$\therefore \displaystyle\int_0^13^xdx = \displaystyle\int_0^1e^{x\ln3}dx$$

$$\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \sin x{ \cos }^{ 3 }xdx } \\ Let\quad \cos x=t\\ \Rightarrow \sin xdx=-dt\\ also,\quad at\quad x=0;t=1\\ and\quad at\quad x=\dfrac { \pi  }{ 2 } ;t=0\\ \therefore \int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \sin x{ \cos }^{ 3 }xdx } =-\int _{ 1 }^{ 0 }{ { t }^{ 3 }dt } \\ ={ -\left[ \dfrac { { t }^{ 4 } }{ 4 }  \right]  }_{ 1 }^{ 0 }\\ =\dfrac { 1 }{ 4 }$$

$$\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\sin x \cos x \ dx}{\cos^4 x+\sin^4 x}$$

consider, $$I = \displaystyle \int_{\pi/3}^{\pi/2} (\tan x + \cot x)^2dx$$

$$\displaystyle \int_0^1\dfrac{2x}{5x^2+1}dx$$
$$x^2=t$$     $$2xdx=dt$$

$$\int _{ 0 }^{ 2 }{ x\sqrt { x+2 } dx }$$
Integration by parts
$$=x\int { \sqrt { x+2 } dx } -\int { \cfrac { dx }{ dx } \left( \int { \sqrt { x+2 } dx }  \right) dx }$$
$$=x\cfrac { { \left( x+2 \right)  }^{ 3/2 } }{ 3/2 } -\int { \cfrac { { \left( x+2 \right)  }^{ 3/2 } }{ 3/2 } dx }$$
$$=2x\cfrac { { \left( x+2 \right)  }^{ 3/2 } }{ 3 } -\cfrac { 2 }{ 3 } \cfrac { { \left( x+2 \right)  }^{ 5/2 } }{ 5/2 } { ] }_{ 0 }^{ 2 }$$
$$=2x\cfrac { { \left( x+2 \right)  }^{ 3/2 } }{ 3 } -\cfrac { 4{ \left( x+2 \right)  }^{ 5/2 } }{ 15 } { ] }_{ 0 }^{ 2 }$$
$$=\cfrac { 4{ \left( 4 \right)  }^{ 3/2 } }{ 3 } -\cfrac { 4{ \left( 4 \right)  }^{ 5/2 } }{ 15 } -0+\cfrac { 4{ \left( 2 \right)  }^{ 5/2 } }{ 15 }$$
$$=\cfrac { 32 }{ 3 } -\cfrac { 4\times 32 }{ 15 } +\cfrac { 16\sqrt { 2 }  }{ 15 }$$
$$=\cfrac { 160 }{ 15 } -\cfrac { 128 }{ 15 } +\cfrac { 16\sqrt { 2 }  }{ 15 }$$
$$=\cfrac { 32 }{ 15 } +\cfrac { 16\sqrt { 2 }  }{ 15 } =\cfrac { 16 }{ 15 } \left( 2+\sqrt { 2 }  \right)$$

$$\int_{2} ^{3} (1 + 2x) dx$$

Property:
$$\displaystyle\int_a^bf(x)dx=\displaystyle\int_a^bf(a+b-x)dx$$

$$I=\displaystyle\int_{-\pi/2}^{\pi/2}\log\left(\dfrac{2-\sin x}{2+\sin x}\right) dx$$     (1)
Using above property:
$$I=\displaystyle\int_{-\pi/2}^{\pi/2}\log\left(\dfrac{2-\sin (-x)}{2+\sin (-x)}\right) dx$$

$$\implies I=\displaystyle\int_{-\pi/2}^{\pi/2}\log\left(\dfrac{2+\sin x}{2-\sin x}\right) dx$$     (2)

Adding (1) and (2):
$$2I=\displaystyle\int_{-\pi/2}^{\pi/2}\log 1 dx = \boxed{0}$$

$$\begin{array}{l} I=\int _{ 0 }^{ \pi  }{ { { \sin   }^{ 3 } }xdx }  \\ =2\int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { { \sin   }^{ 3 } }xdx }  \\ =2\left[ { { { \sin   }^{ 2 } }x\left( { -\cos  x } \right) +\int _{  }^{  }{ \cos  x{ { \sin   }^{ 2 } }xdx }  } \right]  \\ =2\left[ { -{ { \sin   }^{ 2 } }\cos  x+\int _{  }^{  }{ 2\sin { { \cos   }^{ 2 } } dx }  } \right]  \\ =2\int _{  }^{  }{ 2\sin  x{ { \cos   }^{ 2 } }xdx=-\int _{  }^{  }{ 2{ t^{ 2 } }dt }  }  \\ \cos  x=t \\ =-\frac { 2 }{ 3 } { t^{ 3 } }=\frac { { -2 } }{ 3 } { \left( { \cos  x } \right) ^{ 3 } } \\ \frac { { dt } }{ { dx } } =-\sin  x \\ =2\left[ { -{ { \sin   }^{ 2 } }x\cos  x+\frac { { -2 } }{ 3 } { { \left( { { { cosx } } } \right)  }^{ 3 } } } \right] _{ 0 }^{ \frac { \pi  }{ 2 }  } \\ hence, \\ I=2\left[ { 0-\left( { -\frac { 2 }{ 3 }  } \right)  } \right] =\frac { 4 }{ 3 }  \end{array}$$

$$I=\displaystyle \int_0^1\dfrac x{1+x^2}dx$$

$$t=1+x^2\implies dt=2xdx$$

$$x\to 0\to 1\\t\to 1\to 2$$

$$=\dfrac 12\displaystyle \int _1^2\dfrac 1tdt$$

$$=\left.\dfrac 12\log t\right|_1^2=\dfrac 12\log 2$$

$$\int_{0}^{3}{\cfrac{x}{\sqrt{{x}^{2} + 16}} dx}$$

$$I=2\displaystyle \int ^{\pi/2}_{0} \dfrac{\text{sec}^{2} x}{1+3\tan^2 x} d x=\dfrac{2}{3} \int^{\pi/2}_{0} \dfrac{\text{sec}^{2} x}{\dfrac{1}{3}+\tan^2 x}d x=\dfrac{2}{3}\bigg[\text{tan}^{-1}\bigg(\sqrt{3}\tan x\bigg)\bigg]^{\pi/2}_{0}=\dfrac{2\sqrt{3}}{3}\bigg[\dfrac{\pi}{2}-0\bigg]=\dfrac{\pi}{\sqrt{3}}$$

$$I = \dfrac {2}{\pi}\int_{-\dfrac {\pi}{4}^{\pi/4}} \dfrac {dx}{(1 + e^{\sin x})(2 - \cos 2x)} ......(1)$$
by $$a + b - x$$ property
$$I = \dfrac {2}{\pi} \int_{-\dfrac {\pi}{4}}^{\pi/4} \dfrac {dx}{(1 + e^{-sin x})(2 - cos 2x)} = \dfrac {2}{\pi} \int_{-\dfrac {\pi}{4}}^{\pi/4} \dfrac {e^{\sin xdx}}{(1 + e^{\sin x})(2 - \cos 2x)}dx .... (2)$$
adding (1) and (2)
$$2I = \dfrac {2}{\pi} \int_{-\dfrac {\pi}{4}}^{\pi/4} \dfrac {dx}{(1 + e^{-\sin x})(2 - \cos 2x)} = \dfrac {2}{\pi} \int_{-\dfrac {\pi}{4}}^{\pi/4} \dfrac {e^{\sin x}dx}{(1 + e^{\sin x})(2 - \cos 2x)}dx ....(2)$$
put $$\tan x = t, \sec^{2} x dx = dt$$
$$= \dfrac {2}{\pi}\int_{0}^{1} \dfrac {dt}{3t^{2} + 1} = \dfrac {2}{3\pi} \dfrac {1}{\left (\dfrac {1}{\sqrt {3}}\right )} \left (\tan^{-1} \left (\dfrac {t}{1/\sqrt {3}}\right )\right )^{1}_{0} = \dfrac {2}{\sqrt {3}\pi} (\tan^{-1}(\sqrt {3}) - \tan^{-1} (0)) = \dfrac {2}{\sqrt {3}\pi} \left (\dfrac {\pi}{3}\right ) = \dfrac {2}{3\sqrt {3}}$$
Now $$27I^{2} = 27\times \dfrac {4}{27} = 4$$ .

$$I=\displaystyle \int_e^{e^2}\dfrac {1}{\log x}.1\displaystyle \int_e^{e^2}\dfrac {1}{(\log x)^2}dx$$

$$\displaystyle \int_{2}^{3}(2x^2+1)dx$$

$$\displaystyle \int_{1}^{3}(x^2+3x+e^{x})dx$$

Let $$\text{I}=\displaystyle\int \dfrac{1}{5-4\cos x} d x$$

$$\displaystyle \int_{1}^{3}(3x^2+1)dx$$

Let $$\text{I}=\displaystyle\int \dfrac{1}{2+\sin x+\cos x} d x$$

Let $$I=\displaystyle\int \dfrac{1}{x^3}\sin (\log x)d x$$

Let $$I=\displaystyle\int \sqrt{4 x^2-5}\ d x$$

(4) $$I=\int_{0}^{1} \dfrac{\sin ^{-1} \sqrt{x}}{x^{2}-x+1} d x\\$$
$$I=\int_{0}^{1} \dfrac{\sin ^{-1} \sqrt{1-x}}{x^{2}-x+1} d x=\int_{0}^{1} \dfrac{\cos ^{-1} \sqrt{x}}{x^{2}-x+1} d x\\$$
On adding equations (1) and $$(2),$$ we get
$$\begin{array}{l}2 I=\int_{0}^{1} \dfrac{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}{x^{2}-x+1} d x \\=\dfrac{\pi}{2} \int_{0}^{1} \dfrac{d x}{x^{2}-x+1} d x\end{array}\\$$
$$\begin{array}{l}=\dfrac{\pi}{2} \int_{0}^{1} \dfrac{d x}{x^{2}-x+1} d x \\=\dfrac{\pi}{2} \int_{0}^{1} \dfrac{d x}{\left(x-\dfrac{1}{2}\right)^{2}+\left(\dfrac{\sqrt{3}}{2}\right)^{2}} d x \\2 I=\dfrac{\pi}{2} \dfrac{1}{\left(\dfrac{\sqrt{3}}{2}\right)}\left[\tan ^{-1}\left(\dfrac{2 x-1}{\sqrt{3}}\right)\right]_{0}^{1}=\dfrac{\pi^{2}}{3 \sqrt{3}} \\\text { Hence, } I=\dfrac{\pi^{2}}{6 \sqrt{3}}=\dfrac{\pi^{2}}{\sqrt{108}} \equiv \dfrac{\pi^{2}}{\sqrt{n}}\end{array}$$

Let $$I=\int\left(x^{2010}+x^{804}+x^{402}\right)\left(2 x^{1608}+5 x^{402}+10\right)^{1 / 402} d x$$
$$=\int x\left(x^{2009}+x^{508}+x^{401}\right) \cdot\left(2 x^{1608}+5 x^{402}+10\right)^{1 / 402} d x$$
$$=\int\left(x^{2009}+x^{803}+x^{401}\right) \cdot\left(2 x^{2010}+5 x^{304}+10^{402}\right)^{1 / 402} d x$$
Put $$2 x^{2010}+5 x^{804}+10^{x^{402}}=t$$
$$\Rightarrow \quad 4020\left(x^{2009}+x^{803}+x^{401}\right) d x=d t$$
$$\therefore \quad I=\int \dfrac{1}{4020} \cdot(t)^{1 / 402} d t=\dfrac{1}{4020} \cdot \dfrac{t^{1 / 420+1}}{1 / 402+1}$$
$$\quad=\dfrac{1}{4020} \cdot \dfrac{t^{403 / 402}}{403 / 402}$$
$$=\dfrac{1}{4030}\left(2 x^{2010}+5 x^{804}+10^{402}\right)^{403 / 402}$$
$$\therefore a-400=3$$

$$\displaystyle \int^{\beta}_{\alpha} \frac{dx}{(x - \alpha)(\beta - x)}$$
Let, $$I = \displaystyle \int^{\beta}_{\alpha} \frac{dx}{(x - \alpha)(\beta - x)}$$
        $$I = \displaystyle \int^{\beta}_{\alpha} \left ( \frac{A}{x - \alpha} + \frac{B}{\beta - x} \right ) dx$$
On solving, $$A = B = \frac{1}{\beta - \alpha}$$
$$\therefore \frac{1}{\beta - \alpha} \displaystyle \int^{\beta}_{\alpha} \left ( \frac{1}{x - \alpha} + \frac{1}{\beta - x} \right ) dx$$
        $$= \frac{1}{\beta - \alpha} [\log (x - \alpha) + \log (\beta - x)]^{\beta}_{\alpha}$$
        $$= \frac{1}{\beta - \alpha} [\log (\beta - \alpha) + \log (\beta - \beta) - \log (\alpha - \alpha) + \log (\beta - \alpha)]$$
        $$= \frac{1}{\beta - \alpha} [\log (\beta - \alpha) + 0 - 0 + \log (\beta - \alpha)]$$
        $$= \frac{2 \log (\beta - \alpha)}{\beta - \alpha}$$

Let $$I = \displaystyle \int^{\beta}_{\alpha} \frac{dx}{\sqrt{(x - \alpha)(\beta - x)}}, \beta > \alpha$$
and            $$x - \alpha = t^2$$
$$\therefore$$         $$dx = 2t dt$$
when $$x - \alpha$$ , then $$t = 0$$
when $$x - \beta$$ , then $$t = \sqrt{\beta - \alpha}$$
$$I = \displaystyle \int^{\sqrt{\beta - \alpha}}_{0} \frac{2tdt}{\sqrt{t^2 \beta - (\alpha + t^2)}} dx$$
$$I = 2 \displaystyle \int^{\sqrt{\beta - \alpha}}_{0} \frac{dt}{\sqrt{(\beta - \alpha) - t^2}} dx$$
$$= 2 \left [\sin^{-1} \frac{t}{\sqrt{(\beta - \alpha)}}  \right ]^{\sqrt{\beta - \alpha}}_{0}$$
$$= 2 [\sin^{-1} 1 - \sin^{-1} 0] = 2 \left ( \frac{\pi}{2} - 0 \right )$$
$$= \pi$$

$$\displaystyle \int^{1}_{0} \frac{x^3}{\sqrt{1 - x^2}} dx$$
Let $$1 - x^ = t$$
$$\Rightarrow$$     $$-2x dx = dt$$
$$\Rightarrow  xdx = - \frac{1}{2} dt$$
When $$x = 0$$ , then $$t = 1 - 0 = 1$$
When $$x = 1$$ , then $$t = 1 - 1 = 0$$
$$\therefore \displaystyle \int^{1}_{0} \frac{x^3}{\sqrt{1 - x^2}} dx = \int_{0}^{1} \frac{x^2 \cdot x}{\sqrt{1 - x^2}}$$
                                  $$= \int^{0}_{1} \frac{1 - t}{\sqrt t} dt$$
                                      $$= - \frac{1}{2} \int^{0}_{1} (t^{- 1/2} + t^{1/2}) dt$$
                                      $$= - \frac{1}{2} \left [ (2t ^{1/2})^0_1 - \left ( \frac{2t^{3/2}}{3} \right ) ^0_1 \right ]$$
                                      $$= - \frac{1}{3} [(t^{3/2})^0_1] - (t^{1/2})^0_1$$
                                      $$= \frac{1}{3} (0 - 1) - (0 -1)$$
                                      $$= \frac{1}{3} \times (-1) + 1 = 1 - \frac{1}{3} = \frac{2}{3}$$

Let, $$\dfrac {x+3}{x(x+2)} = \dfrac {A}{x} + \dfrac {B}{x+2}$$
$$\Rightarrow \dfrac {x+3}{x(x+2)} = \dfrac {A(x+2)+Bx}{x(x+2)}$$
$$\Rightarrow x+3 = (A+B)x+2A$$
By Comparison,
A+B = 1, 2A = 3, hence A = 3/2
B = 1 - 3/2 = - 1/2

therefore $$\int_{1}^{2} \dfrac {(x+3)}{x(x+2)} dx = \int_{1}^{2} \dfrac {3}{2x} dx - \int_{1}^{2(x+2)} dx$$
$$\Rightarrow \dfrac {3}{2} (log x)_{1}^{2} - \dfrac {1}{2} [log (x+2)]_{1}^{2}$$
$$\Rightarrow \dfrac {3}{2} log 2 - \dfrac {1}{2} log \dfrac{4}{3}$$
$$\Rightarrow \dfrac {1}{2} \left [ log 2^3 - log \dfrac {2^2}{3}\right ]$$
$$\Rightarrow \dfrac {1}{2} log \left ( \dfrac {2^3 \times 3}{2^2}\right ) = \dfrac {1}{2} log 6$$

Hence,
$$\int_{1}^{2} \dfrac {x+3}{x(x+2)} dx = \dfrac {1}{2} log 6$$

Let $$\displaystyle I = \int _{-2}^{2} | 2x + 3 | dx$$
$$= \displaystyle \int_{-2}^{-3/2} - (2x + 3) dx + \int_{-3/2}^{2}(2x + 3) dx$$
          $$[ \therefore x$$ is negative in interval $$\left (-2, -\frac{3}{2} \right )$$
                              $$\therefore |2x + 3| = - (2x + 3)]$$
$$= \displaystyle - \left [ \frac{2x^{2}}{2} + 3x \right ] _{-2}^{-3/2} + \left [\frac{2x^{2}}{2} + 3x \right ] _{-3/2}^{2}$$
$$= (x^{2} + 3x)_{-3/2}^{2} - (x^{2} + 3x)_{-2}^{-3/2}$$
$$= \displaystyle \left [2^{2} + 3 \times 2 - \left (\frac{-3}{2} \right )^{2} - 3 \left (\frac{-3}{2} \right ) \right ]$$
$$= \displaystyle - \left [ \left (\frac{-3}{2} \right ) ^{2} + 3 \left (\frac{-3}{2} \right ) - (-2)^{2} -3( - 2) \right ]$$
$$= \displaystyle \left [ 4 + 6 - \frac{9}{4} + \frac{9}{2} \right ] - \left [ \frac{9}{4} - {9}{2} - 4 + 6 \right ]$$
$$= 4 + 6 - \frac{9}{4} + \frac{9}{2} - \frac{9}{4} + \frac{9}{2} + 4 - 6$$
$$= 8 + 9 - \frac{9}{2} = 8 + \frac{9}{2} = \frac{16 + 9}{2} = \frac{25}{2}$$

$$\int_{1/3}^{1} \dfrac {(x-x^3)^{1/3}}{x^4} dx$$
= $$\int_{1/3}^{1} \dfrac {x \left ( \dfrac {1}{x^2} - 1 \right )^{1/3} dx}{x^4}$$
= $$\int_{1/3}^{1} \dfrac { \left ( \dfrac {1}{x^2} - 1 \right )^{1/3} dx}{x^3}$$

Now, let $$\dfrac {1}{x^2}-1 = t$$ when $$x = 1/3$$ , then $$t=8$$
$$\dfrac {dx}{x^3} = \dfrac {-dt}{2}$$ and $$x=1$$ , then $$t=0$$

Thus, $$\dfrac {-1}{2}\int_{8}^{0} (t)^{1/3} dt$$
= $$\dfrac {1}{2} \int_{0}^{8}  (t)^{1/3} dt$$
= $$\dfrac {1}{2} \times \dfrac {3}{4} \times [t^{4/3}]_{0}^{8}$$
= $$\dfrac {3}{8} \times (8)^{4/3} = 6$$

$$I = \int_{1}^{2} \dfrac {xe^x}{(1+x)^2} dx$$
= $$\int_{1}^{2} \dfrac{(x+1)-1}{(1+x)^2}e^x dx$$
= $$\int_{1}^{2} \left ( \dfrac {1}{1-x} \dfrac {1}{(1+x)^2} \right ) e^x dx$$
= $$\left [ \dfrac {e^x}{1+x} \right ]_{1}^{2}$$
= $$\dfrac {e^2}{1+2} - \dfrac {e^1}{1+1}$$
= $$\dfrac {e^2}{3} - \dfrac {e}{2} = \dfrac {e}{6} (2e - 3)$$

Let $$\displaystyle I = \int_{-2}^{2} |1 - x{2}| dx$$
$$\displaystyle \Rightarrow I = \int_{-2}^{-1} |1 - x^{2}| dx + \int_{-1}^{1} |1 - x^{2}| dx + \int_{1}^{2} |1 - x^{2}| dx$$
$$\displaystyle = - \int_{-2}^{-1} (1 - x^{2}) dx + \int_{-1}^{1} (1 - x^{2}) dx - \int_{1}^{2} |1 - x^{2}| dx$$
$$= \displaystyle - \left [ x - \frac{x^{3}}{3} \right ] _{-2}^{-1} + \left [ x - \frac{x^{3}}{3} \right ] _{-1}^{1} - \left [ x - \frac{x^{2}}{3} \right ] _{1}^{2}$$
$$= \displaystyle - \left [ -1 + \frac{1}{3} - \left (-2 + \frac{8}{3} \right ) \right ] + \left [ 1 - \frac{1}{3} - \left ( -1 + \frac{1}{3} \right ) \right ] - \left [ 2 - \frac{8}{3} - \left ( 1 - {1}{3} \right ) \right ]$$
$$= \displaystyle - \left ( \frac{-4}{3} \right ) + \frac{4}{3} - \left ( \frac{-4}{3} \right )$$
$$\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$$

Let $$\displaystyle I = \int_{1}^{4} f(x) dx$$
      $$\displaystyle I = \int_{1}^{3} (7x + 3) dx + \int_{3}^{4} 8x dx$$
      $$\displaystyle = \left [ 7 \left (\frac{x^{2}}{2} \right ) + 3x \right ] _{1}^{3} + 8 \left [ \frac{x^{2}}{2} \right ] _{3}{4}$$
      $$= \frac{7}{2} (3^{2} - 1^{2}) + 3(3 - 1) + 4(4^{2} - 3^{2})$$
      $$= \frac{7}{2} \times 8 + 6 + 4 \times 7$$
      $$= 28 + 6 + 28 = 62$$      

Let,
$$I = \int_{-2}^{2} \left | 1- x^2 \right | dx$$

$$I=\int _{0} ^{\pi} \dfrac{xsin^{3}x}{1+cos^{2}x}dx$$
$$=\int_{0} ^{\pi} \dfrac{(\pi-x)sin^{3}(\pi-x)}{1+cos^{2}(\pi-x)}dx=\int _{0} ^{\pi} \dfrac{(\pi-x)sin^{3}x}{1+cos^{2}x}dx$$
Hence
$$2I=\int _{0} ^{\pi} \dfrac{\pi sin^{3}x}{1+cos^{2}x}dx$$
Or
$$I=\dfrac{\pi}{2}\int _{0} ^{\pi} \dfrac{sin^{3}x}{1+cos^{2}x}dx$$
Let $$cosx=t$$
$$sinxdx=-dt$$
Hence
$$I=\dfrac{-\pi}{2}\int_{0} ^{-1} \dfrac{(1-t^{2})dt}{1+t^{2}}$$
$$=\dfrac{\pi}{2}\int_{0} ^{-1} \dfrac{(t^{2}-1)dt}{t^{2}+1}$$
$$=\dfrac{\pi}{2}\int_{0} ^{-1} \dfrac{(t^{2}+1-2)dt}{t^{2}+1}$$
$$=\dfrac{\pi}{2}\int_{0} ^{-1} 1-\dfrac{2}{t^{2}+1}dt$$
$$=\dfrac{\pi}{2}[t-2tan^{-1}t]_{0} ^{-1}$$
$$=\dfrac{\pi}{2}[-1-2tan^{-1}(-1)]$$
$$=\dfrac{\pi}{2}[-1-2(\dfrac{-\pi}{4})]$$
$$=\dfrac{\pi}{2}(\dfrac{\pi}{2}-1)$$
$$=\dfrac{\pi^{2}}{4}-\dfrac{\pi}{2}$$

Given : $$\int _{ 0 }^{ 1 }{ x(\tan ^{ -1 }{ x } )^{ 2 }dx }$$

Given : $$\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ \sin { \phi  } \cos { \phi  } \sqrt { a^{ 2 }\sin ^{ 2 }{ \phi  } +{ b }^{ 2 }\cos ^{ 2 }{ \phi  }  }  } d\phi$$

$$\Rightarrow I =\displaystyle \int _{ 0 }^{ \frac { \pi  }{ 2n }  }{ \dfrac { dx }{ 1+\cot ^{ n }{ n\left (\dfrac { \pi  }{ 2n } -x\right) }  }  } =\int _{ 0 }^{ \frac { \pi  }{ 2n }  }{ \dfrac { dx }{ 1+\cot ^{ n }{ \left (\dfrac { \pi  }{ 2 } -nx\right) }  }  }$$

$$\displaystyle \int_{0}^{\frac {1}{2}} \frac {(1 + \sqrt {3})dx}{[(1 + x)^{2} (1 - x)^{6}]^{1/4}}$$

Let us consider the problem:

  $$\int 1/5+3cosx. dx$$

use $$t=tanx/2$$ thus, $$cosx=1-t^2/1+t^2$$ and $$cos^(2)x/2=1/(1+t)^2$$

$$dt/dx=1/2.sec^(2).x/2$$
$$dt/dx=([1+t^2]/2)$$

By substitution, we eliminate cosx etc and get ...
int $$1/5+3cosx. dx = int 2/(8-2t^2). dt$$
=int $$1/(4-t^2).dt$$
= $$1/4.ln([2+t]/[2-t]) + C$$

$$\begin{array}{l} \int _{ 0 }^{ 1 }{ { { \cot   }^{ -1 } }\left( { 1-x+{ x^{ 2 } } } \right) dx } \Rightarrow \int _{ 0 }^{ 1 }{ \frac { 1 }{ { 1-x+{ x^{ 2 } } } } dx }  \\ \int _{ 0 }^{ 1 }{ { { \tan   }^{ -1 } } } \frac { 1 }{ { 1-x\left( { 1-x } \right)  } } dx\Rightarrow \int _{ 0 }^{ 1 }{ { { \tan   }^{ -1 } }\frac { { x+\left( { 1-x } \right)  } }{ { 1-x\left( { 1-x } \right)  } }  } \, \, \, \, \, \, \, \, \, \, \, \, \left[ { { { \tan   }^{ -1 } }\left( { \frac { { a+b } }{ { 1-ab } }  } \right) ={ { \tan   }^{ -1 } }a+{ { \tan   }^{ -1 } }b } \right]  \\ \int _{ 0 }^{ 1 }{ { { \tan   }^{ -1 } }xdx+\int _{ 0 }^{ 1 }{ { { \tan   }^{ -1 } } }  } \left( { 1-x } \right) dx\, \, \, \, \, \, \, \, \, \, \, \, \left[ { \int _{ a }^{ b }{ f\left( x \right) dx=\int _{ a }^{ b }{ f\left( { a+b+x } \right) dx }  }  } \right] \, \, \, \,  \\ \int _{ 0 }^{ 1 }{ { { \tan   }^{ -1 } }xdx+\int _{ 0 }^{ 1 }{ { { \tan   }^{ -1 } }xdx }  } \, \, \, \, \, \, \,  \\ 2\int _{ 0 }^{ 1 }{ 1{ { \tan   }^{ -1 } }xdx } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { un\sin  g\, \, by\, \, parts } \right]  \\ 2\left[ { { { \tan   }^{ -1 } }x\cdot x-\int { \frac { x }{ { 1+{ x^{ 2 } } } } dx }  } \right] \, \, \, \, \, \left[ \begin{array}{l} \, let\, \, 1+{ x^{ 2 } }=t \\ difference= \\ 2xdx=dt \\ xdx=\frac { { dt } }{ 2 }  \end{array} \right]  \\ 2\left[ { x\cdot { { \tan   }^{ -1 } }x-\frac { 1 }{ 2 } \int { \frac { { dt } }{ t }  }  } \right]  \\ 2\left[ { x{ { \tan   }^{ -1 } }x-\frac { 1 }{ 2 } \log  t } \right]  \\ 2\left[ { x{ { \tan   }^{ -1 } }x-\frac { 1 }{ 2 } \log  \left( { 1+{ x^{ 2 } } } \right)  } \right] _{ 0 }^{ 1 } \\ 2\left[ { \frac { \pi  }{ 4 } -\frac { 1 }{ 2 } \log  \sqrt { 2 }  } \right]  \\ =\frac { \pi  }{ 2 } -\log  \sqrt { 2 }  \end{array}$$

(2) I=∫√2+1√2−1(x2+1)2−(x2−1)(x2+1)2dx=∫√2+1√2−1(1−(x2−1)(x2+1)2)dxI=∫2√+12√−1(x2+1)2−(x2−1)(x2+1)2dx=∫2√+12√−1(1−(x2−1)(x2+1)2)dx

$$(0)$$ We have $$J=\int_{-5}^{-4}\left(3-x^{2}\right) \tan \left(3-x^{2}\right) d x\\$$
$$\operatorname{Put}(x+5)=t,$$ we get
$$J=\int_{0}^{1}\left(3-(t-5)^{2}\right) \tan \left(3-(t-5)^{2}\right) d t\\$$
$$=\int_{0}^{1}\left(-22+10 t-t^{2}\right) \tan \left(-22+10 t-t^{2}\right) d t\\$$
Now, $$K=\int_{-2}^{-1}\left(6-6 x+x^{2}\right) \tan \left(6 x-x^{2}-6\right) d x\\$$
Put $$(x+2)=z,\\$$ we get
$$K=\int_{0}^{1}\left(6-6(z-2)+(z-2)^{2}\right) \tan \left(6(z-2)-(z-2)^{2}-6\right) d z\\$$
$$=\int_{0}^{1}\left(22-10 z+z^{2}\right) \tan \left(-22+10 z-z^{2}\right) d z\\$$
Hence, $$(J+K)=0$$

$$\begin{aligned}&\text { Let } I=2 \int_{0}^{\pi / 2} \sin x . \cos x . \tan ^{-1}(\sin x) d x\\&\text { Put } \sin x=z \Rightarrow \cos x d x=d z\\&\text { The limits are, when } x=0, z=\sin 0=0 ; x=\dfrac{\pi}{2}, z=\sin \dfrac{\pi}{2}=1\\&\therefore I=2 \int_{0}^{1} z \tan ^{-1}(z) d z=2\left[\tan ^{-1} z \cdot \dfrac{z^{2}}{2}\right]_{0}^{1}-2 \int_{0}^{1} \dfrac{1}{1+z^{2}} \cdot \dfrac{z^{2}}{2} d z\\&=2\left[\dfrac{\pi}{4} \cdot \dfrac{1}{2}-0\right]-\dfrac{2}{2} \int_{0}^{1} \dfrac{z^{2}}{1+z^{2}} \mathrm{d} \mathbf{z}\\&=\dfrac{\pi}{4}-\int_{0}^{1} \dfrac{1+z^{2}-1}{1+z^{2}} \mathrm{d} \mathbf{z}=\dfrac{\pi}{4}-\int_{0}^{1} \mathrm{d} \mathrm{z}+\int_{0}^{1} \dfrac{\mathrm{d} z}{1+z^{2}}\\&=\dfrac{\pi}{4}-|z|_{0}^{1}+\left[ \tan ^{-1} z\right]_{0}^{1}=\dfrac{\pi}{4}-1+\left[\dfrac{\pi}{4}-0\right]=\dfrac{\pi}{2}-1\end{aligned}$$