How do you find the integral of #sin^3(x) cos^5(x) dx#?

Question

Christopher Allers · Accepted Answer

#cos^8(x)/8-cos^6(x)/6+C#

Recall that through the Pythagorean Identity #sin^2(x)=1-cos^2(x)#.

Thus, #sin^3(x)=sin(x)sin^2(x)=sin(x)(1-cos^2(x))#. Substituting this into the integral we see:

#intsin^3(x)cos^5(x)dx=intsin(x)(1-cos^2(x))cos^5(x)dx#

Distributing just the cosines, this becomes

#=int(cos^5(x)-cos^7(x))sin(x)dx#

Now use the substitution: #u=cos(x)" "=>" "du=-sin(x)dx#

Noting that #sin(x)dx=-du#, the integral becomes:

#=-int(u^5-u^7)du#

Integrating, this becomes

#=-(u^6/6-u^8/8)+C#

Reordering and back-substituting with #u=cos(x)#:

#=cos^8(x)/8-cos^6(x)/6+C#

Note that this integration could have also been done my modifying the cosines like:

#cos^5(x)=cos(x)(cos^2(x))^2=cos(x)(1-sin^2(x))^2#

And then proceeding by expanding and letting #u=sin(x)#.

Charles Arocha · Answer

undefined To integrate sin^3(x) cos^5(x) dx, you can use trigonometric identities and substitution. Start by using the identity sin^2(x) = 1 - cos^2(x) to rewrite sin^3(x) as sin(x)*(1 - cos^2(x)). Then, use substitution where u = cos(x), and du = -sin(x)dx. After substitution, the integral becomes -∫(1 - u^2)u^5 du, which can be integrated term by term. Finally, reverse the substitution to obtain the integral in terms of x.