Explanation
∫10(tan−1x)31+x2dx
∫(tan−1x)31+x2dx=∫(tan−1x)3d(tan−1x)
=(tan−1x)44+c
∫10(tan−1x)31+x2⋅dx=((tan−1x)44+c)10
=[(tan−11)44+c]−[(tan−10)44+1]
=[(π4)44+c]−[0+c]
=π445=π41024
Consider, I=∫π40etanxcos2xdx
∫etanx⋅sec2xdx=∫etanx⋅d(tanx)
=∫etanxd(tanx)=etanx+c
∴∫π40etanxcos2xdx=(etanx+c)π40
=(etanπ4+c)−(etan0+c)
=(e+c)−(1+c)
=e−1
Hence, ∫π40etanxcos2xdx=e−1
∫10tanhxdx ∫tanhxdx=∫ex−e−xex+e−xdx=log(ex+e−x)+c ∫10tanhxdx=log(ex+e−x)+c∫10 =(log(e+1e)+c)−(log(2)+c) =log(e+1e)−log2 =log(e2+12e) ∫10tanhxdx=log(e2+12e)
∫π0tanxsecx+cosxdx
∫tanxsecx+cosxdx=∫d(secx)1+sec2x
=tan−1(secx)+c
∫π0tanxsecx+cosxdx=[tan(secx)+c]|π0
=(tan−1(secπ)+c)−[tan−1(sec0)]
=−π2
∫10xdx(x2+1)2 ∫xdx(x2+1)2=12∫dx2(1+x2)2=−12(11+x2)−1 =−12(1+x2)+c =∫10xdx(x2+1)2=[−12(1+x2)+c]∫10 =[−12(2)+c]−[−12+c] =−14+12=14 ∫10xdx(x2+1)2=14
∫π20dx4cos2x+9sec2x ∫dx4cos2x+9sec2x=∫sec2xdx4+9tan2x=∫d(tanx)4+9tan2x =14tan−1(32tanx)23+c =16tan−1(32tanx)+c ∫π20dx4cos2x+9sec2x=[16tan−1(32tanx)+c]∫π20 =(16tan−1(32tanπ2)+c]−(16tan−1(0)+c) =16×π2=π12
⇒ I=23∫13011−u2du=23[12log1+u1−u]130
To find : ∫a0x−ax+a⋅dx Consider, ∫x−ax+a⋅dx=∫1−2a∫1(x+a)dx =x−2alog(x+a)+c Then, ∫a0x−ax+a⋅dx=[x−2alog(x+a)+c]a0 =(a−2alog(1a)+c)−[0−2alog(a)+c] =a−2alog2 ∫a0x−ax+a⋅dx=a−2alog2
∫10(1−x1+x)dx
∫1−x1+xdx=∫21+xdx−∫1⋅dx
=2log(1+x)−x+c
∫1−x1+xdx=2log(1+x)−x+c
∫101−x1+xdx=(2log(1+x)−x+c)10
=2log(2)−1
=log4−log e
=log(4e)
∫21dx√1+x2=log|x+√x2+a|+c ∫21dx√1+x2=log|x+√x2+1|+c =log|2+√5|−log|1+√2| =log|2+√51+√2| ∫21dx√1+x2=log|2+√51+√2|
=∫1−1dx1+x2 ∫dx1+x2=tan−1x+c ∫1−1dx1+x2=(tan−1x+c)∫1−1 =tan−11−tan−1(−1) =π4−(−π4) =π2 So, ∫1−1dx1+x2=π2
∫21√−x2+3x−2dx
=∫21√−x2+3x−94+94−2dx
=∫21√−(x−32)2+(12)2dx=∫21√(12)2−(x−32)2dx
=(x2√(x−1)(2−x)+(12)22sin−1(x−3212)+c)21
=(x2√(x−1)(2−x)+18sin−1(2x−3)+c)21
=18sin−1(1)−18sin−1(−1)
=18(π2+π8)
=π8
∫∞1[11+x2]dx tan−1x∫∞1 π2−π4=π4
∫π20sin4xcos2xdx ∫π20sinmxcosxdx Im,n=(m−1m+n)(m−3m+n−2)(m−5m+n−4).....I0,norI1,,,,, I4,2=(4−16)(4−34)I0,2 ∫π20cos2xdx=12×π2 36(14)×π4=π32
∫31dx√(x−1)(3−x) ∫31dx√−3+4x−x2=∫31dx√1−(x−2)2 =∫31dx√1−(x−2)2=[sin−1(x−21)+c]31 =sin−1(x−2)31 =sin−1(1)−sin−1(−1) =π2−(−π2) =\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
\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}
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