# Evaluating $\int\frac{\cos^{n-1}\frac{x+a}{2}}{\sin^{n+1}\frac{x-a}{2}}\;dx$

Please help me to evaluate the following integral: $$\int\dfrac{\cos^{n-1}\dfrac{x+a}{2}}{\sin^{n+1}\dfrac{x-a}{2}}\;dx$$

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## 1 Answer

$$\int\frac{\cos^{n-1}\frac{x+a}{2}}{\sin^{n+1}\frac{x-a}{2}}dx=\int\frac{\cos^{n-1}\frac{x+a}{2}}{\sin^{n-1}\frac{x-a}{2}}\cdot\frac{dx}{\sin^2\frac{x-a}{2}}=\left|\frac{\cos\frac{x+a}{2}}{\sin\frac{x-a}{2}}=t\Rightarrow\frac{dx}{\sin^2\frac{x-a}{2}}=-\frac{2dt}{\cos a}\right|=-\frac{2}{\cos a}\int{t^{n-1}dt}=-\frac{2}{\cos a}\cdot\frac{t^n}{n}=-\frac{2}{n\cos a}\cdot\frac{\cos^n\frac{x+a}{2}}{\sin^n\frac{x-a}{2}}+C.$$

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Use double dollar signs (two before and two after the equation) renders the equation in "display mode", which uses larger font and more generous spacing. The result is much easier to read. –  Austin Mohr Sep 5 '12 at 15:49
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