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Question

Evaluate: 10sin1(2x1+x2)dx

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Solution

I=10sin1(2x1+x2)dx
Let x=tanθ
dx=sec2θdθ
2x1+x2=2tanθ1+tan2θ=sin2θ
When x=0θ=0
When x=1θ=π4
=π40sin1(sin2θ)sec2θdθ
=π402θsec2θdθ
Let u=θdu=dθ
dv=sec2θdθv=tanθ
=2[θtanθ]π402π40tanθdθ
=2[π4tanπ40]2[ln|secx|]π40
=π22[lnsecπ4ln|sec0|]
=π22[ln2ln|1|]
=π22[ln20]
=π22ln2
=π22ln212
=π2ln2

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