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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following:
$\int\limits^\pi_0\text{x}\sin\text{x}\cos^2\text{xdx}$
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Answer

Let $\text{I}=\int\limits^\pi_0\text{x}\sin\text{x}\cos^2\text{xdx}\ \ \dots(\text{i})$
$\Rightarrow\ \text{I}=\int\limits^\pi_0(\pi-\text{x})\sin(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$
$\bigg[\because\int\limits^\text{b}_\text{a}\text{f(x)}\text{dx}=\int\limits^\text{b}_\text{a}\text{f(a}+\text{b}-\text{a})\text{dx}\bigg]$
$\Rightarrow\ \text{I}=\int\limits^\pi_0(\pi-\text{x})\sin\text{x}\cos^2\text{x}\text{ dx}$
Adding Eqs. (i) and (ii), we get
$2\text{I}=\int\limits^\pi_0\pi\sin\text{x}\cos^2\text{x dx}$
Now, put $\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0,$ then $\text{t}\rightarrow1$
and $\text{x}\rightarrow\pi,$ then $\text{t}\rightarrow-1$
$\therefore\ 2\text{I}=-\pi\int\limits^{-1}_1\text{t}^2\text{dt}$
$\Rightarrow\ 2\text{I}=-\pi\Big[\frac{\text{t}^3}{3}\Big]^{-1}_1$
$\Rightarrow\ 2\text{I}=-\frac{\pi}{3}[-1-1]=\frac{2\pi}{3}$
$\therefore\ \text{I}=\frac{\pi}{3}$

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