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Question

Evaluate the definite integral π20cos2xdxcos2x+4sin2x

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Solution

Let I=π20cos2xcos2x+4sin2xdx
I=π20cos2xcos2x+4(1cos2x)dx
I=π20cos2xcos2x+44cos2xdx
I=13π2043cos2x443cos2xdx
I=13π2043cos2x43cos2xdx+13π20443cos2xdx
I=13π201dx+13π204sec2x4sec2x3dx
I=13[x]π20+13π204sec2x4(1+tan2x)3dx
I=π6+23π202sec2x1+4tan2xdx .............. (1)
Consider, π202sec2x1+4tan2xdx
Let 2tanx=t2sec2xdx=dt
When x=0,t=0 and when x=π2,t=
π202sec2x1+4tan2xdx=0dt1+t2
=[tan1t]0
=[tan1()tan1(0)]=π2
Therefore, from (1), we obtain
I=π6+23[π2]=π3π6=π6

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