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Exercise 9.10 · Q18

Q.The value of ∫01(sin⁡−1x)2 dx\displaystyle\int_0^1 \left(\sin^{-1}x\right)^2\,dx is

(1) π24−1\dfrac{\pi^2}{4}-1
(2) π24+2\dfrac{\pi^2}{4}+2
(3) π24+1\dfrac{\pi^2}{4}+1
(4) π24−2\dfrac{\pi^2}{4}-2
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The substitution x=sin⁡θx=\sin\theta converts the inverse-sine-squared integrand into a polynomial-times-trig integrand, which Bernoulli's formula (polynomial u=θ2u=\theta^2, easily-integrated v=cos⁡θv=\cos\theta) handles in one pass.

Step 1. Substitute x=sin⁡θx=\sin\theta. Then dx=cos⁡θ dθdx=\cos\theta\,d\theta and sin⁡−1x=θ\sin^{-1}x=\theta. When x=0,θ=0x=0,\theta=0; when x=1,θ=π/2x=1,\theta=\pi/2.

∫01(sin⁡−1x)2 dx=∫0π/2θ2cos⁡θ dθ.\int_0^1(\sin^{-1}x)^2\,dx=\int_0^{\pi/2}\theta^2\cos\theta\,d\theta.

Step 2. Apply Bernoulli's formula with u=θ2u=\theta^2 (polynomial) and v=cos⁡θv=\cos\theta.

u=θ2, u(1)=2θ, u(2)=2, u(3)=0u=\theta^2,\ u^{(1)}=2\theta,\ u^{(2)}=2,\ u^{(3)}=0 (terminates).

v(1)=sin⁡θ, v(2)=−cos⁡θ, v(3)=−sin⁡θv_{(1)}=\sin\theta,\ v_{(2)}=-\cos\theta,\ v_{(3)}=-\sin\theta.

∫θ2cos⁡θ dθ=θ2sin⁡θ−2θ(−cos⁡θ)+2(−sin⁡θ)=θ2sin⁡θ+2θcos⁡θ−2sin⁡θ.\int\theta^2\cos\theta\,d\theta=\theta^2\sin\theta-2\theta(-\cos\theta)+2(-\sin\theta)=\theta^2\sin\theta+2\theta\cos\theta-2\sin\theta.

Step 3. Evaluate at θ=π/2\theta=\pi/2. …

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