Question
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_0\cos^5\text{x dx}$

Answer

$\int^\limits{\frac{\pi}{2}}_0\cos^5\text{x dx}=\int^\limits{\frac{\pi}{2}}_0\big(1-\sin^2\text{x}\big)^2\cos\text{x dx}$
Let $\sin\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\cos\text{x dx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$=\int^\limits{\frac{\pi}{2}}_0\big(1-\sin^2\text{x}\big)^2\cos\text{x dx}$
$=\int^\limits{1}_0\big(1-\text{t}^2\big)^2\text{ dt}$
$=\int^\limits{1}_0\big(1-2\text{t}^2+\text{t}^4\big)\text{dt}$
$=\Big[\text{t}-\frac{2}{3}\text{t}^3+\frac{\text{t}^5}{5}\Big]^1_0$
$=1-\frac{2}{3}+\frac{1}{5}$
$=\frac{8}{15}$

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