Evaluate int_0^{2 pi} cos m x cos n x , dx + | vedclass

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Question

(B) Let $I = \int_0^{2 \pi} \cos m x \cos n x \, dx + \int_{-\pi}^\pi \sin m x \cos n x \, dx$. Consider the second integral $J = \int_{-\pi}^\pi \sin

Vedclass · Accepted Answer

$\pi$,if $m = n 
eq 0$

Vedclass · Answer

(B) Let $I = \int_0^{2 \pi} \cos m x \cos n x \, dx + \int_{-\pi}^\pi \sin m x \cos n x \, dx$. Consider the second integral $J = \int_{-\pi}^\pi \sin m x \cos n x \, dx$. Since $\sin m(-x) \cos n(-x) = -\sin m x \cos n x$,the integrand is an odd function. Thus,$J = 0$. Now,consider $I = \int_0^{2 \pi} \cos m x \cos n x \, dx$. Using the identity $\cos A \cos B = \frac{1}{2} [\cos(A-B) + \cos(A+B)]$,we have: $I = \frac{1}{2} \int_0^{2 \pi} [\cos((m-n)x) + \cos((m+n)x)] \, dx$. If $m 
eq n$ and $m, n \in \mathbb{Z}$,then $\int_0^{2 \pi} \cos(kx) \, dx = 0$ for $k 
eq 0$. Thus,$I = 0$ if $m 
eq n$. If $m = n 
eq 0$,then $I = \int_0^{2 \pi} \cos^2(mx) \, dx = \int_0^{2 \pi} \frac{1 + \cos(2mx)}{2} \, dx = \frac{1}{2} [x + \frac{\sin(2mx)}{2m}]_0^{2 \pi} = \frac{1}{2} (2 \pi) = \pi$. Therefore,the value is $\pi$ if $m = n 
eq 0$.

Evaluate $\int_0^{2 \pi} \cos m x \cos n x \, dx + \int_{-\pi}^\pi \sin m x \cos n x \, dx$ for $m, n \in \mathbb{Z}$.

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