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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.6
Integrate the functions in Exercises 134 to 155.
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Question : 143 of 271
Marks: +1, -0
(sin⁡−1x)2(\sin^{-1}x)^2
Solution:  
Let I=∫(sin⁡−1x)2dxI=\int(\sin^{-1}x)^2dx
Put sin⁡−1x=θ\sin^{-1}x=\theta ⇒x=sin⁡θ\Rightarrow x=\sin\theta ⇒dx=cos⁡θdθ\Rightarrow dx=\cos\theta d\theta
∴I=∫θ2cos⁡θdθ\therefore I=\int\theta^2\cos\theta d\theta
=θ2∫cos⁡θdθ−∫(ddθθ2⋅∫cos⁡θdθ)dθ=\theta^2\int\cos\theta d\theta-\int\left(\frac{d}{d\theta}\theta^2\cdot\int\cos\theta d\theta\right)d\theta
=θ2sin⁡θ−∫2θ(sin⁡θ)dθ=\theta^2\sin\theta-\int2\theta(\sin\theta)d\theta
=θ2sin⁡θ−2∫θsin⁡θdθ+C=\theta^2\sin\theta-2\int\theta\sin\theta d\theta+C
=θ2sin⁡θ−2[θ(−cos⁡θ)−∫1⋅(−cos⁡θ)dθ]+C=\theta^2\sin\theta-2\left[\theta(-\cos\theta)-\int1\cdot(-\cos\theta)d\theta\right]+C
=θ2sin⁡θ+2θcos⁡θ−2∫cos⁡θdθ+C=\theta^2\sin\theta+2\theta\cos\theta-2\int\cos\theta d\theta+C
=θ2sin⁡θ+2θ1−sin⁡2θ−2sin⁡θ+C=\theta^2\sin\theta+2\theta\sqrt{1-\sin^2\theta}-2\sin\theta+C
=x(sin⁡−1x)2+2sin⁡−1x1−x2−2x+C=x(\sin^{-1}x)^2+2\sin^{-1}x\sqrt{1-x^2}-2x+C
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