Question: solve the following integral
∫cos4 θ sin θ dθ.
You can do it by applying the integration by parts.
Let me recall the general formula for Integration by Part
∫u(x)v'(x)dx = u(x)v(x)- ∫u'(x)v(x)dx let u(x) be cos4θ and let v'(x) be sinθ
∫cos4θ sinθ dθ = cos4θ (-cosθ) - ∫(-4cos3θ sinθ)(-cosθ) dθ = -cos5θ -4∫cos4θ sinθ dθ
So on the right side we find the same integral we want solve.
We can rewrite only the first and the last part,
∫cos4θ sinθ dθ = -cos5θ -4∫cos4θ sinθ dθ,
now, like an equation, we move -4∫cos4θ sinθ dθ on the left side,
5∫cos4θ sinθ dθ = -cos5θ
and finally, dividing by 5, we have the solution
∫cos4θ sinθ dθ =(-cos5θ )/5 + C