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Question: \[\int_{}^{}{\sin^{–1}(3x–4x^{3})dx =}\]...

∫sin⁡–1(3x–4x3)dx=\int_{}^{}{\sin^{–1}(3x–4x^{3})dx =}

A

x sin–1x + 1–x2\sqrt{1–x^{2}} + c

B

x sin–1x – 1–x2\sqrt{1–x^{2}} + c

C

2[x sin–1x +1–x2\sqrt{1–x^{2}}] + c

D

3[x sin–1x +1–x2\sqrt{1–x^{2}}] + c

Answer

3$$x sin–1x +1–x2\sqrt{1–x^{2}}$$ + c

Explanation

Solution

Put x = sinq

dx = cosq dq

\ ∫sin⁡−1\int \sin ^ { - 1 }(3sinq – 4 sin3q) cosq dq

= ∫sin⁡−1\int \sin ^ { - 1 } (sin 3q) . cos dq = 3∫θ.cos⁡θ.dθ3\int_{}^{}{\theta.\cos\theta.d\theta}

= 3 [q∫cos⁡θdθ\int_{}^{}{\cos\theta d\theta}–∫{ddθ(θ).∫cos⁡θdθ}dθ]\int_{}^{}{\{\frac{d}{d\theta}(\theta).\int_{}^{}{\cos\theta d\theta\} d\theta}\rbrack}

= 3 [θsin⁡θ–∫sin⁡θdθ]\lbrack\theta\sin\theta –\int_{}^{}{\sin\theta d\theta\rbrack} = 3[q sin q + cosq] + c

= 3[x sin–1x + 1–x2\sqrt{1–x^{2}}] + c