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By integrating by parts twice, prove that InI_n as defined in the first equality below for positive integers nn has the value given in the second equality:

In=0π/2sinnθcosθdθ=nsin(nπ/2)n21I_n=\int_0^{\pi / 2} \sin n \theta \cos \theta d \theta=\frac{n-\sin (n \pi / 2)}{n^2-1} \text {. }

Solution

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Answered 2 years ago
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Given:0π2sinnθcosθdθ\textbf{Given}: \int _{0}^{\frac{\pi}{2}} \sin n\theta \cos\theta d\theta

Solving by parts:

uvdx=uvdx(dudxvdx)dx\int u v dx= u\int v dx- \int \left(\dfrac{du}{dx} \int v dx \right)dx

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