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Question

Evaluate sin2x0sin1ttd+cos2x0cos1tdtx

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Solution

Consider sin2x0sin1tdt
Let t=sin2θdt=2sinθcosθdθ=sin2θdθ
When t=0θ=0
When t=sin2xsin2θ=sin2xθ=x
sin2x0sin1tdt
=x0sin1sin2θsin2θdθ
=x0sin1sinθsin2θdθ
=x0θsin2θdθ ...........(1)
Consider cos2x0cos1tdt
Let t=cos2θdt=2sinθcosθdθ=sin2θdθ
When t=0θ=π2
When t=cos2xcos2θ=cos2xθ=x
xπ2cos1tdt
=xπ2cos1cos2θsin2θdθ
=xπ2cos1cosθsin2θdθ
=xπ2θsin2θdθ
=π2xθsin2θdθ ...........(2)
Using baf(x)dx+cbf(x)dx=caf(x)dx we add (1) and (2)
x0θsin2θdθ+π2xθsin2θdθ
=π20θsin2θdθ
Let u=θdu=dθ
dv=sin2θv=cos2θ2
=[θcos2θ2]π20cos2θ2dθ
=[π40]π20cos2θ2dθ
=π4+14[sin2θ]π20
=π4+14[00]π20
=π4

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