Indefinite Integration 1 Question 3
4. The value of $\int _0[\sin 2 x(1+\cos 3 x)] d x$, where $[t]$ denotes the greatest integer function, is
(2019 Main, 10 April I)
(a) $-\pi$
(b) $2 \pi$
(c) $-2 \pi$
(d) $\pi$
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Solution:
- Given integral
$$ \begin{aligned} I= & \int _0^{2 \pi}[\sin 2 x \cdot(1+\cos 3 x)] d x \\ & =\int _0^{\pi}[\sin 2 x \cdot(1+\cos 3 x)] d x \\ & \quad \quad \quad \quad+\int _{\pi}^{2 \pi}[\sin 2 x \cdot(1+\cos 3 x)] d x \\ & =I _1+I _2 \text { (let) } \end{aligned} $$
Now, $\quad I _2=\int _{\pi}^{2 \pi}[\sin 2 x \cdot(1+\cos 3 x)] d x$
let $2 \pi-x=t$, upper limit $t=0$ and lower limit $t=\pi$
and $\quad d x=-d t$
So, $\quad I _2=-\int _{\pi}^{0}[-\sin 2 x \cdot(1+\cos 3 x)] d x$
$=\int _0^{\pi}[-\sin 2 x \cdot(1+\cos 3 x)] d x$
$\begin{aligned} \therefore \quad I & =\int _0^{\pi}[\sin 2 x \cdot(1+\cos 3 x)] d x \ & +\int _0^{\pi}[-\sin 2 x \cdot(1+\cos 3 x)] d x\end{aligned}$
[from Eqs. (i) and (ii)]
$\left.=\int _0^{\pi}(-1) d x\right][\because[x]+[-x]=-1, x \notin$ Integer $]$
$=-\pi$
