Search any question & find its solution
Question:
Answered & Verified by Expert
$\int_0^{2 \pi} \cos m x \cos n x d x+\int_{-\pi}^\pi \sin m x \cos n x d x=$
Options:
Solution:
1054 Upvotes
Verified Answer
The correct answer is:
$\pi$ if $m=n, m, n \in \mathbf{Z}$
since $f(-x)=\sin m(-x) \cos x(-x)$
$\begin{aligned} & f(-x)=-\sin x \cos x \\ & =-f(x)\end{aligned}$
$\therefore \mathrm{f}(\mathrm{x})$ is odd function
$\begin{aligned} & \therefore \int_{-\pi}^\pi \sin m x \cdot \cos n x d x=0 \\ & \therefore I=\int_0^{2 \pi} \cos m x \cdot \cos n x d x=0\end{aligned}$
$2 \mathrm{t} \mathrm{m}=\mathrm{n}$
$\begin{aligned} & =\int_0^{2 \pi} \cos ^2 x d x \\ & I=4 \int_0^{\pi / 2} \cos ^2 x d x \quad \ldots \text { (i) }\left[\because \cos ^2 x \text { is over function }\right] \\ & I=4 \int_0^{\pi / 2} \cos ^2\left(\frac{\pi}{2}-x\right) d x\end{aligned}$
$I=4 \int_0^{\pi / 2} \sin ^2 x d x$...(ii)
Adding (i) and (ii)
$\begin{aligned} & 2 I=4 \int_0^{\pi / 2} 1 \mathrm{dx}=4 \cdot \frac{\pi}{2}=2 \pi \\ & \therefore \mathrm{I}=\pi\end{aligned}$
$\begin{aligned} & f(-x)=-\sin x \cos x \\ & =-f(x)\end{aligned}$
$\therefore \mathrm{f}(\mathrm{x})$ is odd function
$\begin{aligned} & \therefore \int_{-\pi}^\pi \sin m x \cdot \cos n x d x=0 \\ & \therefore I=\int_0^{2 \pi} \cos m x \cdot \cos n x d x=0\end{aligned}$
$2 \mathrm{t} \mathrm{m}=\mathrm{n}$
$\begin{aligned} & =\int_0^{2 \pi} \cos ^2 x d x \\ & I=4 \int_0^{\pi / 2} \cos ^2 x d x \quad \ldots \text { (i) }\left[\because \cos ^2 x \text { is over function }\right] \\ & I=4 \int_0^{\pi / 2} \cos ^2\left(\frac{\pi}{2}-x\right) d x\end{aligned}$
$I=4 \int_0^{\pi / 2} \sin ^2 x d x$...(ii)
Adding (i) and (ii)
$\begin{aligned} & 2 I=4 \int_0^{\pi / 2} 1 \mathrm{dx}=4 \cdot \frac{\pi}{2}=2 \pi \\ & \therefore \mathrm{I}=\pi\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.