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Question

Evaluate: π0xdxa2cos2x+b2sin2x

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Solution

Let I=π0xdxa2cos2x+b2sin2x ...(i)

I=π0πxa2cos2(πx)+b2sin2(πx) ......[ By using a0f(x)dx=a0f(ax)dx]

I=π0πxa2cos2x+b2sin2x(ii)

Adding (i) and (ii), we get

2I=π0πa2cos2x+b2sin2xdx

I=π2π0dxa2cos2x+b2sin2x

Dividing numerator and denominator by cos2x, we get

I=π2π0sec2xdxa2+b2tan2x

I=ππ20sec2xdxa2+b2tan2x ......[ By using 2a0f(x)dx=2a0f(x)dx]

Putting btanx=tbsec2 x dx=dt

The limits are, when x=0,t=0 and x=π2,t

I=πb0dta2+t2=πb1atan1ta]0I=πab(tan1tan10)=πabπ2I=π22ab

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