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Question

If I=dx(1+sinx)4=A4992(17u7+35u5+u3+u)+C, where u=12sinx1+sinx then A is equal to

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Solution

Let I=1(sinx+1)4dx
Substituting t=tanx2

dt=12sec2x2dx

sinx=2tt2+1,cosx=1t2t2+1,dx=2dtt2+1
I=2(t2+1)(2tt2+1+1)4dt=2(t2+1)3(t+1)8dt

=8((t2)2t2+2)6+3((t2)2t2+2)4+3((t2)2t2+2)2+1((1t2)(t2+1)(2t+1)2)dt
Let (t2)2t2+2=u12sinxsinx+1=u
I=8(u6+3u4+3u2+1)du=8(u77+35u4+u3+u)

Comparing with, I=dx(1+sinx)4=A4992(17u7+35u5+u3+u)+C
Therefore, A=624

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