Trig Integral: int sin^2(x)*cos^2(x) dx

[des4ij]

Hey, I had a question on the following:

{ sin^2(x)*cos^2(x) dx

The way I am looking at this is simplifying it to sin^2(x)-sin^4(x) then doing it some real long way. Is there a simpler method. Thank You.

 

[soroban]

Hello, des4ij!

You're expected to be familiar with these Double-angle identities:

. . \(\displaystyle \L\sin^2x\:=\:\frac{1\,-\,\cos2x}{2}\) . . . . \(\displaystyle \L\cos^2x\:=\:\frac{1\,+\,\cos2x}{2}\)

\(\displaystyle \L\int\sin^2x\cos^2x\,dx\)

We have: \(\displaystyle \L\:\int\left(\frac{1\,-\cos2x}{2}\right)\left(\frac{1\,+\,\cos2x}{2}\right)\,dx \;=\;\frac{1}{4}\int\left(1\,-\,\cos^22x\right)\,dx\)

. . \(\displaystyle \L=\;\frac{1}{4}\int\sin^22x\,dx \;=\;\frac{1}{4}\int\frac{1\,-\,\cos4x}{2}\,dx \;=\;\frac{1}{8}\int\left(1\,-\,\cos4x\right)\,dx\)

. . \(\displaystyle \L=\;\frac{1}{8}\left(x\,-\,\frac{1}{4}\sin4x\right)\,+\,C\)

 

[des4ij]

thx... another way...

hey also my teacher just gave me a hint telling me to change it to (sin(X)cos(X))^2... i get how u did it... but any ideas how u get about from (sin(X)cos(X))^2 instead... its cool though... thank you...