Evaluate the following integrals: ^ _0x ^4x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}$

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Answer

Let $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^4(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^4\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin\text{x}\cos^4\text{x dx}$
$=\pi\int\limits^{\pi}_0\sin\text{x}\cos^4\text{x dx}$
Let $\cos\text{x}=\text{t},$ Then $-\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1,\text{x}=\pi,\text{t}=-1$
Therefore, $2\text{I}=-\pi\int\limits^{-1}_1\text{t}^4\text{ dt}$
$=\pi\int\limits^{1}_{-1}\text{t}^4\text{ dt}$
$=\pi\Big[\frac{\text{t}^5}{5}\Big]^{1}_{-1}$
$=\frac{\pi}{5}+\frac{\pi}{5}$
$=\frac{2\pi}{5}$
Hence, $\text{I}=\frac{\pi}{5}$

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