how do you solve the integration of sin x^5?

Answer 1

(sin x)^5 = (sin x)^4 * sin x = [(sin x)^2]^2 * sin x

= [ 1- (cosx)^2 ] ^2 *sin x = - [ 1- u^2]^2 du

(u= cos x). this can then be integrated by first multiplying out the squared term.

this method works in principle for any odd power of sine, but of course you could end up raising 1-u^2 to a huge power.

OR......
integration by parts gives the "reduction formula" that appears in tables of integrals. (this is the formula that appears in the post below mine)

Answer 2

This is one way, but I do not know whether it's the best way.
integ sin^n (x) = -sin^(n-1)(x) cos(x) / n + (n-1) /n integ sin^(n-2) (x)
let n=5
You will get again integ sin^3(x) in the above relation. Repeat it for sin^3(x), finally , you'll get integ sin(x) which you know is -cos(x).

Answer 3

ask for forgiveness, repent and avoid the occasions of sin.