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Question

Solve:
sin2x9sin4xdx

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Solution

We have,

sin2x9sin4xdx

Now,

sin2x32(sin2x)2dx

Let

sin2x=t

ddxsin2x=dt

2sinxcosxdx=dt

sin2xdx=dt

Now,

dt32t2

Using that,

1a2x2dx=sin1xa

dt32t2=sin1t3+C

dt32t2=sin1sin2x3+C

Hence, this is the answer.

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