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Let's calculate the integration

$\int_{0}^{\infty}\frac{r^{n-1}}{(1+r^2)^{(n+1)/2}}dr.$

then let $r=\tan\theta$,

according to my book, its result is

$\int_{0}^{\pi/2}\sin^{n-1}\theta d\theta.$

but my calculation is

$\int_{0}^{\pi/2}\cos^{2}\theta \sin^{n-1}\theta d\theta.$

Is my book wrong?

1 Answers 1

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The book is right. When you make the substitution $r=\tan\theta$, you have $dr=\sec^2\theta\,d\theta$.

So our integral is equal to $\int_0^{\pi/2} \frac{\tan^{n-1}\theta}{\sec^{n+1}\theta}\sec^2\theta\,d\theta,$ which simplifies to the book's expression.