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Question

Integrate sinxcosxdx.


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Solution

Calculate the given expression:

Given, sinxcosxdx

Let, I=sinxcosxdx

Let, sinx=t

Differentiating both sides with respect to x,

cosx=dtdxcosxdx=dt

I=tdt[sinx=t,cosxdx=dt]I=t22+C[xn=xn+1n+1]I=sin2x2+C

Therefore,sin(x)cos(x)dx=sin2(x)2+C.


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