What is the integral of #int sin(x)^3 * cos(x) dx#?

Question

7198 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-05T09:32:03

#=(sin^4(x))/(4)+C#

#int_# #sin^3(x)*cos(x) dx#

We can use substitution to remove #cos(x)#. So, let's use #sin(x)# as our source.

#u=sin(x)#

Which then means that we will get,

#(du)/(dx)=cos(x)#

Finding #dx# will give,

#dx= 1/cos(x) * du#

Now replacing the original integral with the substitution,

#int_# #u^3*cos(x) * 1/cos(x) du#

We can cancel out #cos(x)# here,

#int_# #u^3 du#

#= 1/(3+1)u^(3+1)+C = 1/4 u^4+C#

Now setting in for #u#,

#=sin(x)^4/4+C = sin^4(x)/4 + C#