Question #2d98d

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Answer 1 · 2017-12-19T08:40:35

The answer is 0.

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 :)