We want to evaluate:
$$ I = \int_{-\pi}^{\pi} \sin(mx) \cos(nx) dx $$
Using the product-to-sum trigonometric identity:
$$ \sin(mx) \cos(nx) = \frac{1}{2}[\sin((m+n)x) + \sin((m-n)x)] $$
The integral becomes:
$$ I = \frac{1}{2} \int_{-\pi}^{\pi} \sin((m+n)x) dx + \frac{1}{2} \int_{-\pi}^{\pi} \sin((m-n)x) dx $$
Since the sine function is an odd function, integrating any sine function over a symmetric interval \( [-\pi, \pi] \) yields exactly \( 0 \).
Thus, evaluating gives \( 0 + 0 = 0 \). This holds true for all integers \( m \) and \( n \).
We want to evaluate:
$$ I = \int_{-\pi}^{\pi} \sin(mx) \sin(nx) dx $$
Using the product-to-sum trigonometric identity:
$$ \sin(mx) \sin(nx) = \frac{1}{2}[\cos((m-n)x) - \cos((m+n)x)] $$
Subcase A: \(m \neq n\)
Integrating gives: \( \frac{1}{2} \left[ \frac{\sin((m-n)x)}{m-n} - \frac{\sin((m+n)x)}{m+n} \right] \) from \(-\pi\) to \(\pi\). Since \(\sin(k\pi) = 0\) for any integer \(k\), the terms evaluate to 0. Thus, \(I = 0\).
Subcase B: \(m = n \neq 0\)
If \(m = n\), then \(\cos((m-n)x) = \cos(0) = 1\). The integral becomes: $$ \frac{1}{2} \int_{-\pi}^{\pi} [1 - \cos(2mx)] dx $$
Which evaluates to: $$ \frac{1}{2} \left[ x - \frac{\sin(2mx)}{2m} \right]_{-\pi}^{\pi} = \frac{1}{2} ( (\pi - 0) - (-\pi - 0) ) = \frac{1}{2}(2\pi) = \pi $$
We want to evaluate:
$$ I = \int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx $$
Using the product-to-sum trigonometric identity:
$$ \cos(mx) \cos(nx) = \frac{1}{2}[\cos((m-n)x) + \cos((m+n)x)] $$
Subcase A: \( m \neq n \)
Similar to Sine, integrating yields sine terms which evaluate to 0 at bounds \(-\pi\) and \(\pi\). Thus, \( I = 0 \).
Subcase B: \( m = n \neq 0 \)
If \( m = n \), the integral is $$ \frac{1}{2} \int_{-\pi}^{\pi} [1 + \cos(2mx)] dx $$ which similarly evaluates to exactly \( \pi \).