If the primitive of f(x)=13sin⁡x+sin3⁡xis equal to 16log⁡t−1t+1+112log⁡2+t2−t+C then

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t=cos⁡x

t=tan⁡x/2

t=2cos⁡x

t=sin⁡x

answer is A.

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Detailed Solution

∫dx3sin⁡x+sin3⁡x=∫sin⁡xdx3+sin2⁡xsin2⁡x=∫sin⁡xdx4−cos2⁡x1−cos2⁡x=−∫dt4−t21−t2=−13∫1t2−4−1t2−1 (t=cos⁡x)=16log⁡t−1t+1+112log⁡2+t2−t+C

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