CBSE Questions for Class 12 Commerce Maths Integrals Quiz 2 - MCQExams.com

Consider, $$I=\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{e^{\tan x}}{\cos ^2  x}dx$$

$$\displaystyle \int e^{\tan x}\cdot \sec ^2 x  dx =\int e^{\tan x}\cdot d({\tan x})$$

$$=\int e^{\tan x} d(\tan  x )=e^{\tan x}+c$$

$$\therefore \displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{e^{\tan x}}{\cos ^2 x}dx=\left ( e^{\tan x}+ c \right )_{0}^{\frac{\pi}{4}}$$

$$=\left ( e^{\tan \frac{\pi}{4}}+c\right ) -\left ( e^{\tan 0}+c \right )$$

$$=(e+c)-(1+c)$$

$$=e-1$$

Hence, $$\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{e^{\tan x}}{\cos ^2  x}dx=e-1$$

$$\int_{0}^{1} tan  h x   dx$$
$$\int tan  h  x  dx=\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}} dx =log (e^x+e^{-x})+c$$
$$\int_{0}^{1} tan  h x  dx = log (e^x+e^{-x})+c \int_{0}^{1}$$
$$=\left ( log (e+\dfrac{1}{e}) +c \right ) - \left ( log (2)+c \right )$$
$$=log (e+\dfrac{1}{e})-log 2$$
$$=log (\dfrac{e}{2}+\dfrac{1}{2e})$$
$$\int_{0}^{1}tan  hx  dx=log (\dfrac{e}{2}+\dfrac{1}{2e})$$

$$\int_{0}^{\pi}\dfrac{\tan  x}{\sec  x+\cos  x}dx$$

$$\int \dfrac{\tan  x}{\sec  x+\cos  x}dx=\int \dfrac{d  (\sec  x)}{1+\sec ^2 x}$$

$$=\tan ^{-1}(\sec  x)+c$$

$$\int_{0}^{\pi}\dfrac{\tan  x}{\sec  x+\cos  x}dx=\left [ \tan  (\sec  x)+c \right ] |_{0}^{\pi}$$

$$=(\tan ^{-1}(\sec  \pi)+c)- [ \tan ^{-1}(\sec  0) ]$$

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

$$\int_{0}^{1}\dfrac{x  dx}{(x^2+1)^2}$$
$$\int \dfrac{x  dx}{(x^2+1)^2} =\dfrac{1}{2}\int \dfrac{dx^2}{(1+x^2)^2}=-\dfrac{1}{2}\left ( \dfrac{1}{1+x^2} \right )^{-1}$$
$$=-\dfrac{1}{2}(1+x^2)^{+c}$$
$$=\int_{0}^{1}\dfrac{x  dx}{(x^2+1)^2}=\left [ -\dfrac{1}{2} (1+x^2)+c \right ] \int_{0}^{1}$$
$$=\left [ -\dfrac{1}{2} (2)+c \right ] - \left [ -\dfrac{1}{2}+c \right ]$$
$$=-\dfrac{1}{4}+\dfrac{1}{2} =\dfrac{1}{4}$$
$$\int_{0}^{1}\dfrac{x  dx}{(x^2+1)^2}=\dfrac{1}{4}$$

$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{4  cos^2 x  +  9  sec^2  x}$$
$$\int \dfrac{dx}{4  cos^2  x  + 9 sec^2  x}=\int \dfrac{sec^2 x  dx} {4+9  tan^2  x}=\int \dfrac{d(tan  x)}{4+9  tan^2  x}$$
$$=\dfrac{1}{4} tan^{-1}\left ( \dfrac{3}{2} tan  x \right )  \dfrac{2}{3} +c$$
$$=\dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan x \right )  +c$$
$$\int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{4  cos^2 x  +  9  sec^2  x}=\left [ \dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan  x \right ) +c \right ]\int_{0}^{\dfrac{\pi}{2}}$$
$$=\left (\dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan \dfrac{\pi}{2} \right ) +c \right ]- \left ( \dfrac{1}{6}tan^{-1}(0)+c \right )$$
$$=\dfrac{1}{6}\times \dfrac{\pi}{2}=\dfrac{\pi}{12}$$

$$\Rightarrow$$ $$\displaystyle I=\frac { 2 }{ 3 } \int _{ 0 }^{ \frac { 1 }{ 3 }  }{ \cfrac { 1 }{ 1-{ u }^{ 2 } } du } =\cfrac { 2 }{ 3 } { \left[ \cfrac { 1 }{ 2 } \log { \cfrac { 1+u }{ 1-u }  }  \right]  }_{ 0 }^{ \frac { 1 }{ 3 }  }$$

To find : $$\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a} \cdot  dx$$
Consider, $$\displaystyle \int \dfrac{x-a}{x+a} \cdot  dx=\int 1 - 2a \int \dfrac{1}{(x+a)}dx$$
$$=x-2a \log (x+a)+c$$
Then, $$\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a} \cdot  dx=\left[x-2a \log (x+a)+c\right]_{0}^{a}$$
$$=(a-2a \log (1a)+c)- [0-2a \log (a)+c]$$
$$=a-2a \log 2$$
$$\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a}\cdot dx= a-2a \log 2$$

$$\displaystyle \int_{0}^{1} \left ( \dfrac{1-x}{1+x} \right )  dx$$

$$\displaystyle \int  \dfrac{1-x}{1+x} dx = \displaystyle \int \dfrac{2}{1+x}dx -\displaystyle \int1 \cdot  dx$$

$$=2\log (1+x) - x + c$$

$$\displaystyle \int \dfrac{1-x }{1+x}dx = 2 \log (1+x)- x+c$$

$$\displaystyle \int_{0}^{1}\dfrac{1-x}{1+x}dx= \left(2 \log (1+x)-x+c\right)_{0}^{1}$$

$$=2 \log (2)-1$$

$$=\log 4- \log \ e$$

$$=\log \left(\dfrac{4}{e}\right)$$

$$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log |x+\sqrt{x^2+a}|+c$$
$$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log |x+\sqrt{x^2+1}|+c$$
$$=\log |2+\sqrt{5}|-\log |1+\sqrt{2}|$$
$$=\log \left | \dfrac{2+\sqrt{5}}{1+\sqrt{2}} \right |$$
$$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log \left | \dfrac{2+\sqrt{5}}{1+\sqrt{2}} \right |$$

$$=\int_{-1}^{1}\dfrac{dx}{1+x^2}$$
$$\int \dfrac{dx}{1+x^2}=tan^{-1}x+c$$
$$\int_{-1}^{1}\dfrac{dx}{1+x^2}=(tan^{-1}x+c)\int_{-1}^{1}$$
$$=tan^{-1}1-tan^{-1}(-1)$$
$$=\dfrac{\pi}{4}-(\dfrac{-\pi}{4})$$
$$=\dfrac{\pi}{2}$$
So,
$$\int_{-1}^{1}\dfrac{dx}{1+x^2}=\dfrac{\pi}{2}$$

$$\displaystyle \int_{1}^{2}\sqrt{-x^2+3x-2}dx$$

$$=\displaystyle \int_{1}^{2}\sqrt{-x^2+3x-\dfrac{9}{4}+\dfrac{9}{4}-2} dx$$

$$=\displaystyle \int_{1}^{2}\sqrt{-\left (x-\dfrac{3}{2}\right)^2+\left (\dfrac{1}{2}\right)^2}dx=\int_{1}^{2}\sqrt{\left (\dfrac{1}{2}\right)^2-\left (x-\dfrac{3}{2}\right)^2}dx$$

$$=\left ( \dfrac{x}{2}\sqrt{(x-1)(2-x)+\dfrac{(\dfrac{1}{2})^2}{2}}\sin^{-1}\left ( \dfrac{x-\dfrac{3}{2}}{\dfrac{1}{2}} \right )+c \right )_{1}^{2}$$

$$=\left ( \dfrac{x}{2}\sqrt{(x-1)(2-x)}+\dfrac{1}{8}\sin^{-1}(2x-3)+c \right )_{1}^{2}$$

$$=\dfrac{1}{8}\sin^{-1}(1)-\dfrac{1}{8}\sin^{-1}(-1)$$

$$=\dfrac{1}{8}\left ( \dfrac{\pi}{2}+\dfrac{\pi}{8} \right )$$

$$=\dfrac{\pi}{8}$$

$$\int_{1}^{\infty }\left [ \dfrac{1}{1+x^2} \right ]  dx$$
$$tan^{-1}x \int_{1}^{\infty }$$
$$\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$$

$$\int_{0}^{\dfrac{\pi}{2}} sin ^4  x  cos^2  x  dx$$
$$\int_{0}^{\dfrac{\pi}{2}}sin ^m  x  cos ^x   dx$$
$$I_{m, n}=\left (\dfrac{m-1}{m+n} \right )\left (\dfrac{m-3}{m+n-2} \right )\left (\dfrac{m-5}{m+n-4} \right ) .....  I_0, n    or   I_1,,,,,$$
$$I_{4,2}=\left (\dfrac{4-1}{6} \right )\left (\dfrac{4-3}{4} \right )I_{0,2}$$
$$\int_{0}^{\dfrac{\pi}{2}}cos^2x   dx=\dfrac{1}{2}\times \dfrac{\pi}{2}$$
$$\dfrac{3}{6}\left (\dfrac{1}{4} \right )\times \dfrac{\pi}{4}=\dfrac{\pi}{32}$$

$$\int_{1}^{3}\dfrac{dx}{\sqrt{(x-1)(3-x)}}$$
$$\int_{1}^{3}\dfrac{dx}{\sqrt{-3+4x-x^2}}=\int_{1}^{3}\dfrac{dx}{\sqrt{1-(x-2)^2}}$$
$$=\int_{1}^{3}\dfrac{dx}{\sqrt{1-(x-2)^2}}=\left [ sin^{-1}\left (\dfrac{x-2}{1}\right )+c\right ]_1^3$$
$$=sin^{-1}(x-2)_{1}^{3}$$
$$=sin^{-1}(1)-sin^{-1}(-1)$$
$$=\dfrac{\pi}{2}-\left ( \dfrac{-\pi}{2}\right )$$
$$=\pi$$

$$\int_{0}^{\infty}\dfrac{dx}{(x+\sqrt{x^2+1})^5}=\int_{0}^{\infty}\dfrac{(x-\sqrt{x^2+1})^5}{-1}dx$$
$$=\int_{0}^{\infty}(\sqrt{x^2+1}-1)^5  dx$$
$$x=Sin  h  \theta$$
$$\int_{0}^{\infty}(cos  h  \theta-  sin  h  \theta)^5  cos  h  \theta  d\theta$$
$$\int_{0}^{\infty}(e^{-x})^5\left (\dfrac{e^x+e^{-x}}{2} \right ) d\theta$$
$$\int_{0}^{+\infty}\dfrac{e^{-4x}+e^{-6x}}{2}  dx=\dfrac{1}{2}\left [ \left (\dfrac{-1}{4}\right )[0-1]+\left (\dfrac{-1}{6} \right )[0-1] \right ]$$
$$=\dfrac{1}{2}\left ( \dfrac{1}{4}+\dfrac{1}{6}\right )=\dfrac{1}{2}\left (\dfrac{1+4}{24}\right )$$
$$=\dfrac{5}{24}$$

$$\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{sin   2x}{cos^4  x  +  sin^4  x}dx=\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{d  sin^2  x}{(1-sin^2  x)^2+sin^4x}$$

$$=\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{d  sin^2  x}{(2  sin^2  x-2 sin^2  x +\dfrac{1}{2}+\dfrac{1}{2})}$$

$$sin^2  x=t$$ $$\displaystyle \int_{0}^{\dfrac{1}{2}}\dfrac{dt}{(\sqrt{2}t-\dfrac{1}{\sqrt{2}})^2+ (\dfrac{1}{\sqrt{2}})^2}=\dfrac{1}{\sqrt{2}}\dfrac{1}{(\dfrac{1}{\sqrt{2}})}tan^{-1}\left [ \dfrac{\left (\sqrt{2}t- \dfrac{1}{\sqrt{2}} \right )}{\dfrac{1}{\sqrt{2}}} \right ]_{0}^{\dfrac{1}{2}}$$

$$=tan^{-1}(2t-1)_{0}^{{1}/{2}}$$

$$=tan^{-1}(0)-tan^{-1}(-1)$$

$$=0-\left ( \dfrac{-\pi}{4} \right )$$

$$=\dfrac{\pi}{4}$$

$$\int_{a}^{0}\dfrac{x^5dx}{\sqrt{a^2-x^2}}x=a  sin \theta$$
$$\int_{o}^{a}\dfrac{a^5sin ^5\theta}{a   cos  \theta}a cos \theta d  \theta$$
$$\int_{0}^{\dfrac{\pi}{2}}a^5  sin^5\theta  d  \theta=a^5\int_{0}^{\dfrac{\pi}{2}}sin ^5\theta  d  \theta$$
$$=a^5\left ( \dfrac{4}{5}\right )\left ( \dfrac{2}{3}\right )$$
$$=\dfrac{8a^5}{15}$$

$$\displaystyle \int_{0}^{1} cos^{-1}\left ( \dfrac{1-x^2}{1+x^2} \right )$$

$$x=tan \theta$$

$$dx =sec^2 \theta   \cdot  d  \theta$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{4}}cos^{-1}(cos   2\theta)cos^2\theta   d\theta$$

$$\displaystyle \int_{0}^{\dfrac{\pi}{4}}2  \theta   sec^2 \theta   d \theta =2\displaystyle \int_{0}^{\dfrac{\pi}{4}}\theta   sec^2 \theta   d  \theta$$

$$2 [ \theta  tan \theta  +  log  |cos \theta| ]_{0}^{\dfrac{\pi}{4}}$$

$$2 \left [ (\dfrac{\pi}{4}-0)+ log (\dfrac{1}{\sqrt{2}})\right ]$$

$$\dfrac{\pi}{2}- 2 log (\sqrt{2})=\dfrac{\pi}{2}-\dfrac{2}{2}log (2)$$

$$=\dfrac{\pi}{2}-log  2$$

$$\displaystyle \int_{\frac {1}{3}}^{1} \dfrac {(x – x^{3})^{\frac {1}{3}}}{x^{4}} dx = \int_{\frac {1}{3}}^{1} \frac {(x^{3})^{\frac {1}{3}} \left (\dfrac {1}{x^{2}} – 1\right )^{\frac {1}{3}}}{x^{4}} dx$$

$$\displaystyle = \int_{\dfrac {1}{3}}^{1} \dfrac {\left (\dfrac {1}{x^{2}} – 1\right )^{\frac {1}{3}}}{x^{3}} dx$$            (let $$\dfrac {1}{x^{2}} – 1 = t$$ )

$$\displaystyle = \int_{8}^{0} \dfrac{t^{\frac{1}{3}}}{-2} dt$$     $$\therefore \dfrac {-2}{x^{3}} dx = dt$$

$$\int_{0}^{\infty}\left ( \dfrac{a^x}{c^x}-\dfrac{b^x}{c^x} \right ) dx$$
$$=\int_{0}^{\infty}\left ( \dfrac{a}{c} \right )^xdx-\int_{0}^{\infty}\left ( \dfrac{b}{c}\right )^x dx$$
$$=\dfrac{(\dfrac{a}{c})^x}{log (\dfrac{a}{c})}\int_{0}^{\infty}-\dfrac{(\dfrac{b}{c})^x}{log (\dfrac{b}{c})}\int_{0}^{\infty}$$
$$=\left [ 1-\dfrac{1}{log(\dfrac{a}{c})} \right ] - \left [ 1-\dfrac{1}{log(\dfrac{b}{c})}\right ]$$
$$=\dfrac{1}{log(\dfrac{b}{c})}-\dfrac{1}{log(\dfrac{a}{c})}$$

$$\displaystyle \int_{0}^{1}e^x\left ( \dfrac{1}{(x=+1)}-\dfrac{1}{(x+1)^2}\right ) dx$$

It is in the form of

$$\displaystyle \int  e^x \left [ f(x)+f^1 (x) \right ]  dx=e^x f(x)+c$$

$$=\left [ \dfrac{e^x}{(1+x)}+c \right ] _{0}^{1}$$

$$=\dfrac{e}{2}-1$$

$$=\dfrac{e}{2}-1$$

$$\int_{\sqrt{2}}^{x}\dfrac{dx}{x\sqrt{x^2-1}}=\dfrac{\pi}{2}$$
$$=\left ( sin^{-1}x+c \right ) \int_{\sqrt{2}}^{x}=\dfrac{\pi}{2}$$
$$\left ( sec^{-1}x - sec^{-1}\sqrt{2} \right ) =\dfrac{\pi}{2}$$
$$sec^{-1}x- \dfrac{\pi}{4}=\dfrac{\pi}{12}$$
$$sec^{-1}x =\dfrac{\pi}{12}+\dfrac{3\pi}{12}$$
$$sec^{-1}=\dfrac{\pi}{3}$$
$$x=sec  \dfrac{\pi}{3}$$
$$=2$$

$$\int_{0}^{1}\dfrac{dx}{x^2+x  cos_\alpha+1}$$
$$\int_{0}^{1}\dfrac{dx}{(x+cos  \alpha)^2}+ sin^2    \alpha$$
$$\left ( \dfrac{1}{sin  \alpha} \left [ tan^{-1} \left ( \dfrac{x+ cos  \alpha}{sin  \alpha} \right ) \right ]+c \right )_{0}^{1}$$
$$=\dfrac{1}{sin  \alpha} \left [ tan^{-1} \left ( \dfrac{1+cos  \alpha}{sin  \alpha} \right )  \right ]-\dfrac{1}{sin  \alpha}tan^{-1}\left ( \dfrac{cos  \alpha}{sin  \alpha} \right )$$
$$=\dfrac{1}{sin  \alpha}\left [ tan^{-1} (cot  \dfrac{\alpha}{2})-tan^{-1}(cot  \alpha) \right ]$$
$$=\dfrac{1}{sin \alpha} \left ( \alpha -\dfrac{\alpha}{2} \right ) =\dfrac{\alpha}{2} sin  \alpha$$

$$\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4  sin^2  x}dx =\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4  cos^2  x}dx$$
$$\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4  cos^2x}dx =\int_{0}^{\dfrac{\pi}{4}} \dfrac{sec^2x}{(sec^2x + 4)}dx$$
$$=\int_{0}^{\dfrac{\pi}{4}} \dfrac{d(tan   x)}{5 + tan^2  x} dx= \left [ \dfrac{1}{\sqrt{5}} tan^{-1}\left ( \dfrac{tan  x}{\sqrt{5}} \right ) + c \right ]_{0}^{\dfrac{\pi}{2}}$$
$$=\dfrac{1}{\sqrt{5}}(\dfrac{\pi}{2})$$
$$=\dfrac{\pi}{2\sqrt{5}}$$

$$\int_{0}^{16}\dfrac{dx}{\sqrt{x+9}-\sqrt{x}}$$
$$\int_{0}^{16}\dfrac{\sqrt{x}+9+\sqrt{x}}{9}dx$$
$$=\dfrac{1}{19}\left [ \int_{0}^{16} \sqrt{x+9}dx + \int_{0}^{16}\sqrt{x}dx \right ]$$
$$=\dfrac{1}{19}\times \dfrac{2}{3}\left [ (x+a)^{\dfrac{3}{2}}\int_{0}^{16}+ x^{\dfrac{3}{2}} \int_{0}^{16} \right ]$$
$$=\dfrac{2}{27}\left [ (25)^{\dfrac{3}{2}} - (9)^{\dfrac{3}{2}}+ (16)^{\dfrac{3}{2}} \right ]$$
$$=\dfrac{2}{27}\left [ 125-27+64 \right ]$$
$$=\dfrac{2}{27}\left [ 189-27 \right ] =\dfrac{2}{27}\left [ 162 \right ]$$
$$=12$$

$$\int_{0}^{\dfrac{\pi}{4}}\dfrac{sec^2  x \sqrt{tan  x}}{tan  x}dx$$
dividing  and multiplying by $$sec^2 x$$
$$=\int_{0}^{\dfrac{\pi}{4}}\dfrac{d  tan  x}{\sqrt{tan  x}}$$
$$=(2\sqrt{tanx}+c)\int_{0}^{\dfrac{\pi}{4}}$$
$$=2$$

$$\int_{1}^{2}\dfrac{dx}{(x-1)^2+(\sqrt{3})^2}=\left [ \dfrac{1}{\sqrt{3}} tan^{-1} \left ( \dfrac{x-1}{\sqrt{3}}\right ) +c \right ] \int_{1}^{2}$$
$$=\dfrac{1}{\sqrt{3}}tan^{-1}\left ( \dfrac{1}{\sqrt{3}} \right )$$
$$=\dfrac{1}{\sqrt{3}}\cdot \dfrac{\pi}{6}$$
$$=\dfrac{\pi}{6\sqrt{3}}$$

$$x=t^2$$
$$dx=2t   dt$$
$$=\int_{0}^{1}\left ( \dfrac{t^2}{1+t} \right )  2t \cdot  dt$$
$$=2\int_{0}^{1}\dfrac{t^3}{1+t}dt =2\int_{0}^{1}\dfrac{t^3+1}{t+1}dt -2\int_{0}^{1}\dfrac{1}{t+1}dt$$
$$=2\left [ \dfrac{t^3}{3}\dfrac{t^2}{2}+t \right] \int_{0}^{1} -2 log (1+1) \int_{0}^{1}$$
$$=2\left [ \dfrac{1}{3} -\dfrac{1}{2} +1 \right ] -2 log (2)$$
$$=2 \left [ \dfrac{2-3+6}{6} \right ] -2  log (2)$$
$$=\dfrac{5}{3}-log  4$$

$$\int_{0}^{\pi}\dfrac{dx}{3+2  sin  x+  cos  x}$$
$$=\int_{0}^{\pi}\dfrac{dx(1+tan^2 \dfrac{x}{2})}{3(1+tan^2 \dfrac{x}{2}) + 2 (2 tan \dfrac{x}{2}) + (1- tan^2 \dfrac{x}{2})}$$
$$=\int_{0}^{\pi}\dfrac{sec^2\dfrac{x}{2} dx}{2 tan^2 \dfrac{x}{2}+ 4 tan \dfrac{x}{2}+4}$$
$$=\int_{0}^{\pi}\dfrac{2 d (tan^2 \dfrac{x}{2})}{(\sqrt{2}tan \dfrac{x}{2}+\sqrt{2})^2+(\sqrt{2})^2}$$
$$=\int_{0}^{\pi}\dfrac{d (tan^2 \dfrac{x}{2})}{(\sqrt{2}tan \dfrac{x}{2}+\sqrt{2})^2+(\sqrt{2})^2}$$
$$=\int_{0}^{\pi}\dfrac{d(tan^2 \dfrac{x}{2})}{(tan \dfrac{x}{2}+1)^2} =tna^{-1}(tan \dfrac{x}{2}+1)\int_{0}^{\pi}$$
$$=\dfrac{\pi}{2} -\dfrac{\pi}{4}$$
$$=\dfrac{\pi}{4}$$