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If the primitive of f(x)=13sinx+sin3x

is equal to 16logt1t+1+112log2+t2t+C then

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a
t=cos⁡x
b
t=tan⁡x/2
c
t=2cos⁡x
d
t=sin⁡x

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detailed solution

Correct option is A

∫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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