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

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Answer 1 · 2016-05-15T19:58:31

#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)#.