I need help on integration problems

Question

∫cos⁴ θ sin θ dθ

Answer 1

Use the substitution u=cosθ. Then cos⁴θ=u⁴ and du=-sinθ dθ and the integral becomes

 

∫cos⁴ θ sin θ dθ = -∫u⁴du = -1/5 u⁵ + C = -1/5 cos⁵ θ + C

Answer 2

 

∫cos⁴ θ sin θ dθ

= ∫cos⁴ θ dcos θ , mental substitution

= (1/5)cos⁵ θ + C

Answer 3

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

 

Solution:

∫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