How do you prove the given formula where m and n are positive integers?
#int_-pi^pisin(mx)*cos(nx)dx = 0#

Question

Elijah Abram · Accepted Answer

Let #f(x)=sin(mx)cos(nx)#.

Let's think about

#f(-x)=sin(-mx)cos(-nx)#

since #sin(mx)# is odd, and #cos(nx)# is even,

#=-sin(mx)cos(nx)=-f(x)#,

which means that #f(x)# is an odd function.

Recall that if #f(x)# is odd,

#int_{-a}^a f(x)dx=0#.

Hence,

#int_{-pi}^{pi}sin(mx)cos(nx) dx=0#

Ezekiel Alexander · Answer

undefined To prove the given formula, we can use the properties of definite integrals and trigonometric identities. Let's proceed with the proof:

$$ \int_{-\pi}^{\pi} \sin(mx) \cos(nx) \, dx $$

Using the identity $ \sin(A) \cos(B) = \frac{1}{2} [\sin(A + B) + \sin(A - B)] $, we rewrite the integrand:

$$ \frac{1}{2} \int_{-\pi}^{\pi} [\sin((m+n)x) + \sin((m-n)x)] \, dx $$

Now, we integrate each term separately:

For the term $ \sin((m+n)x) $:
$$ \frac{1}{2} \left[ -\frac{\cos((m+n)x)}{m+n} \right]_{-\pi}^{\pi} = 0 $$

For the term $ \sin((m-n)x) $:
$$ \frac{1}{2} \left[ -\frac{\cos((m-n)x)}{m-n} \right]_{-\pi}^{\pi} = 0 $$

Therefore, each term evaluates to zero when integrated over the interval $[- \pi, \pi]$. Since the integral of the sum of two functions is the sum of their integrals, the overall integral is also zero.

Hence, we have proved the given formula:

$$ \int_{-\pi}^{\pi} \sin(mx) \cos(nx) \, dx = 0 $$

How do you prove the given formula where m and n are positive integers? #int_-pi^pisin(mx)*cos(nx)dx = 0#