Evaluate the following integrals: ^ 2 _ 0 ( x 2 + x 2 )^n dx

Question

Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$

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Answer

Let $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^{\text{n}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^{\text{n}-1}}\text{ dx}$
Put $\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}=\text{z}$
$\therefore\ \Big(-\sin\frac{\text{x}}{2}\times\frac{1}{2}+\cos\frac{\text{x}}{2}\times\frac{1}{2}\Big)\text{dx}=\text{dz}$
$\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)\text{dx}=2\text{dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\sqrt{2}$ $\Big(\text{z}=\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\Big)$
$\therefore\ \text{I}=2\int^\limits{\sqrt{2}}_{1}\frac{\text{dz}}{\text{z}^{\text{n}-1}}$
$=2\times\Big[\frac{\text{z}^{\text{n}-1}}{2-\text{n}}\Big]^{\sqrt{2}}_1$
$=\frac{2}{(2-\text{n})}\Big[\big(\sqrt{2}\big)^{2-\text{n}}-1\Big]$
$=\frac{2}{(2-\text{n})}\bigg(2^{\frac{2}{(2-\text{n})}}-1\bigg)$
$=\frac{2}{(2-\text{n})}\Big(2^{1-\frac{\text{n}}{2}}-1\Big)$

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