Integral question please help.

What is ∫(cos(x))^5 dx? I don't really need the answer. I just need the method. How would you go about solving it?

∫(cos(x))^5 dx
for this we use a trig substitution. we start by peeling off a cosx:

∫(cos^4x)cosxdx

since cos^2x = 1-sin^2x we have:
cos^4x = (1-sin^2x)^2 and can rewrite the integral as:
∫(1-sin^2x)^2(cosx)dx

now we have a function and its derivative as the integrand. that is what we needed.
let u=sinx
du=cosxdx

substituting we get:
∫(1-u^2)^2du

multiplying out we get:
∫(1-2u^2+u^4)du

now we can integrate:
u - ⅔u³ + ⅕u⁵ + c
sinx - ⅔(sinx)³ + ⅕(sinx)⁵ + c

Whenever you have cos(x) or sin(x) raised to an odd exponent, take one of the factors out:
cos(x)^4 cos(x) dx

Now you see that you have the derivative of sin(x): cos(x) dx
So put t = sin(x).
You can express cos(x)^4 in terms of sine (using sin^2(x) + cos^2(x) = 1). This is possible only when you have even exponent on cos(x), that's why you should initially have odd exponent.

You'll finally get a polynomial in t.

If the exponent is initially even, then you need other methods, like half angle formulas etc.

Hope this helps.

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