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Question

Evaluating π20cos2xdx1+3sin2x

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Solution

Let I = π20cos2xdx1+3 sin2x I=π20sec2xdxsec4x+3sec2x tan2x on Dividing Nr & Dr by cos4x

I=pi20sec2xdxsec2 x(1+tan2 x+3 tan2 x) I=pi20sec2xdx(1+tan2 x)(1+4 tan2 x)

Put tan x = t sec2xdx = dt. When x = 0 Rightarrow t = 0 & when x = π2t=

So I = 0dt(1+t2)(1+4 t2) I=130(11+t241+4 t2)dt

I=13[tan1 t4 tan1(2t)2]0 I=13[(π22×π2)(00)]=π6.


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