Proof of equation with integrals

[kostas]

Homework Statement

Prove that
$$\int cos^mxdx = \frac{cos^{m-1}x sinx}{m} + \frac{m-1}{m}\int cos^{m-2}xdx$$

Homework Equations

$$\int f(x)g'(x) = f(x)g(x) - \int g(x)f'(x) dx$$

The Attempt at a Solution

Going through any of the integrals provides constants that seem to be a problem in proving the above.
Starting from the left side and applying the integral product rule (see 2. -- not sure how the rule is called in English!) we get:

$$\int cos^mxdx = \int cos^{m-1}x \cdot cosx dx$$
With $$f(x) = cos^{m-1}x$$, $$g'(x) = cosx$$, $$g(x) = sinx$$, $$f'(x) = (m-1)cos^{m-2}x$$ we get:

$$cos^{m-1}x \cdot sinx - \int sinx(m-1)cos^{m-2}x dx = cos^{m-1}x \cdot sinx - (m-1)\int cos^{m-2}x \cdot sinx dx$$

With $$f(x) = cos^{m-2}x$$, $$g'(x) = sinx$$, $$g(x) = -cosx$$, $$f'(x) = (m-2)cos^{m-3}x$$ we get:

$$cos^{m-1}x \cdot sinx - (m-1)\left( cos^{m-2}x (-cosx) - \int (-cosx)(m-2)cos^{m-3}x dx \right) =$$

$$cos^{m-1}x \cdot sinx - (m-1)\left( -cos^{m-1}x + (m-2)\int cos^{m-2}x dx \right) =$$

Any ideas on how to continue?

 

[Dick]

In your first parts integration you should have f'(x)=(m-1)*cos(x)^(m-2)*(-sin(x)). You forgot to use the chain rule. That gives you a sin(x)^2 in the parts integral. Change it to 1-cos(x)^2 and rearrange.

 

[kostas]

In your first parts integration you should have f'(x)=(m-1)*cos(x)^(m-2)*(-sin(x)). You forgot to use the chain rule. That gives you a sin(x)^2 in the parts integral. Change it to 1-cos(x)^2 and rearrange.

Ah..so true :) Thank you!