Question
Evaluate the following integrals:$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}$
✓
Answer
Let $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^2\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\pi}_0(\pi-\text{x}+\text{x})\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=\pi\int\limits^{\pi}_0\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=-\pi\int\limits^{\pi}_0\cos^2\text{x}(-\sin\text{x})\text{dx}$
$\Rightarrow2\text{I}=-\pi\Big[\frac{\cos^3\text{x}}{3}\Big]^{\pi}_0$ $\Bigg[\int\big[\text{f(x)}\big]^{\text{n}}\text{f}'(\text{x})\text{dx}=\frac{\big[\text{f(x)}\big]^{\text{n}+1}}{\text{n}+1}+\text{C}\Bigg]$
$\Rightarrow2\text{I}=-\frac{\pi}{3}\big(\cos^3\text{x}-\cos^20\big)$
$\Rightarrow2\text{I}=-\frac{\pi}{3}(-1-1)=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^2\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\pi}_0(\pi-\text{x}+\text{x})\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=\pi\int\limits^{\pi}_0\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=-\pi\int\limits^{\pi}_0\cos^2\text{x}(-\sin\text{x})\text{dx}$
$\Rightarrow2\text{I}=-\pi\Big[\frac{\cos^3\text{x}}{3}\Big]^{\pi}_0$ $\Bigg[\int\big[\text{f(x)}\big]^{\text{n}}\text{f}'(\text{x})\text{dx}=\frac{\big[\text{f(x)}\big]^{\text{n}+1}}{\text{n}+1}+\text{C}\Bigg]$
$\Rightarrow2\text{I}=-\frac{\pi}{3}\big(\cos^3\text{x}-\cos^20\big)$
$\Rightarrow2\text{I}=-\frac{\pi}{3}(-1-1)=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
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