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∫(sin⁻¹(3x−4x³)−cos⁻¹(4x³−3x))dx | JEE

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Question
The value of the definite integral ∫−1/21/2(sin⁡−1(3x−4x3)−cos⁡−1(4x3−3x))dx\displaystyle\int_{-1/2}^{1/2}\Big(\sin^{-1}(3x-4x^{3})-\cos^{-1}(4x^{3}-3x)\Big)dx is
A00
B−π2-\dfrac{\pi}{2}correct
C7π2\dfrac{7\pi}{2}
Dπ2\dfrac{\pi}{2}
Solution
Step 1: Identify the triple-angle forms on the interval. For x∈(−12,12)x\in\left(-\dfrac12,\dfrac12\right), write x=sin⁡θx=\sin\theta with θ∈(−π6,π6)\theta\in\left(-\dfrac{\pi}{6},\dfrac{\pi}{6}\right); then 3x−4x3=sin⁡3θ3x-4x^{3}=\sin 3\theta with 3θ∈(−π2,π2)3\theta\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right), so
sin⁡−1(3x−4x3)=3θ=3sin⁡−1x.\sin^{-1}(3x-4x^{3})=3\theta=3\sin^{-1}x.
 Step 2: For the second term, 4x3−3x=−(3x−4x3)=−sin⁡3θ=sin⁡(−3θ)4x^{3}-3x=-(3x-4x^{3})=-\sin 3\theta=\sin(-3\theta), so it equals cos⁡ ⁣(π2+3θ)\cos\!\left(\dfrac{\pi}{2}+3\theta\right). Since π2+3θ∈(0,π)\dfrac{\pi}{2}+3\theta\in\left(0,\pi\right) lies in the principal range of cos⁡−1\cos^{-1},
cos⁡−1(4x3−3x)=π2+3θ=π2+3sin⁡−1x.\cos^{-1}(4x^{3}-3x)=\dfrac{\pi}{2}+3\theta=\dfrac{\pi}{2}+3\sin^{-1}x.
 Step 3: Subtract the two:
sin⁡−1(3x−4x3)−cos⁡−1(4x3−3x)=3sin⁡−1x−(π2+3sin⁡−1x)=−π2.\sin^{-1}(3x-4x^{3})-\cos^{-1}(4x^{3}-3x)=3\sin^{-1}x-\left(\dfrac{\pi}{2}+3\sin^{-1}x\right)=-\dfrac{\pi}{2}.
 The integrand is the constant −π2-\dfrac{\pi}{2}. Step 4: Integrate the constant over an interval of length 11:
∫−1/21/2(−π2)dx=−π2⋅1=−π2.\int_{-1/2}^{1/2}\left(-\dfrac{\pi}{2}\right)dx=-\dfrac{\pi}{2}\cdot 1=-\dfrac{\pi}{2}.
 Correct answer: (2)
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