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I need help on integration problems

∫cosθ sin θ dθ

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Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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
Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
∫cosθ sin θ dθ
= ∫cosθ dcos θ , mental substitution
= (1/5)cos5 θ + C


dcos θ = sin θ dθ
Giovanna C. | Master’s Degree in Math & 15 years' teaching experienceMaster’s Degree in Math & 15 years' teac...
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