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Question
Evaluate the following : `int sin θ.log (cos θ).dθ`
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Solution
Let I = `int sin θ.log (cos θ).dθ`
= `int log(cosθ).sinθ dθ`
Put cos θ = t
∴ – sin θ dθ = dt
∴ sin θ dθ = – dt
∴ I = `int logt.(- dt)`
= `- int (logt).1dt`
= `-[(log t) int 1dt - int {d/dt (log t) int 1 dt }dt]`
= `-[(logt)t - int1/t.t dt]`
= `- t logt + int 1 dt`
= – t log t + t + c
= – cos θ . log (cos θ) + cos θ + c
= – cos θ [log (cos θ) – 1] + c.
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