Question

Question #2d98d

Answer 1

The answer is 0.

Explanation:

Since we can pull out `a^3`, we can focus on

`int_0^pi sin^3(2phi) sin(phi) dphi`.

Using the double-angle formula for the sine, i.e., `sin(2phi) = 2 sin(phi) cos(phi)`, this is the same as

`8 int_0^pi sin^4(phi) cos^3(phi) dphi`.

Now we use the identity `cos^2(phi) = 1 - sin^2(phi)`, and arrive at

`8 int_0^pi sin^4(phi) (1 - sin^2(phi)) cos(phi) dphi`.

At this point we identify an inner function `sin(phi)`, i.e., `x = sin(phi)` and `dx = cos(phi) dphi`. Therefore, we get

`8 int_sin(0)^sin(phi) x^4 (1 - x^2) dx = 8 int_0^0 x^4 (1 - x^2) dx = 0.`

Hope this helps :)