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Question

Find the value of the integral 1+2cosx2+cosx2dx from 0 to π2.


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Solution

Solve the above integral.

The given integral is 1+2cosx2+cosx2dx

Assume that, I=0π21+2cosx2+cosx2dx

I=0π21+2cosx2+cosx2dxI=0π2sin2x+cos2x+2cosx2+cosx2dx[sin2x+cos2x=1]I=0π2sin2x+cosxcosx+22+cosx2dxI=0π2cosx2+cosxdx+0π2sin2x2+cosx2dx

Assume that, I1=0π2cosxcosx+2dx

I1=1cosx+20π20π2cosxdx-0π2d1cosx+2dxcosxdxdxI1=sinxcosx+20π2-0π2--sinx2+cosx2sinxdxI1=sinxcosx+20π2-0π2sinx2+cosx2sinxdx

So, I=I1+0π2sin2x2+cosx2dx

I=sinxcosx+20π2-0π2sinx2+cosx2sinxdx+0π2sinx2+cosx2sinxdxI=sinxcosx+20π2I=sinπ2cosπ2+2-sin0cos0+2I=10+2-01+2I=12

Hence, the required final answer is 12.


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