trig integration: int sin^3x*cos^2x dx

[calchere]

Problem:

int sin^3x*cos^2x dx

int (sin^2x*sinx)*cos^2x dx

int [(1-cos^2x)(sinx)]*1/2(1+cos2x) dx

int (sinx-sinxcos^2x)*(1/2*1/2cos2x) dx

The answer is: 1/5cos^5x-1/3cos^3x+C

I am lost. I'm not sure i'm even on the right track.

 

[tkhunny]

The answer suggests getting there with Integration by Parts.

There are also Reduction Formulas for this sort of thing.

 

[soroban]

Re: trig integration

Herllo, calchere!

\(\displaystyle \L \int\)$\displaystyle \sin^3x\cdot\cos^2x\,dx$

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

\(\displaystyle \L\int\)$\displaystyle (1\,-\,\cos^2x)(\sin x)]\cdot\underbrace{\,\frac{1}{2}(1\,+\,\cos2x)\,}\, dx$
. . . . . . . . . . . . . . . . . . . No!

Don't introduce 2x into the problem
. . unless you can change everything to 2x

The answer is: $\displaystyle \:\frac{1}{5}\cos^5x \,-\,\frac{1}{3}\cos^3x\,+\,C$

Your third statement should have been:
. . \(\displaystyle \L\:\int\)\(\displaystyle \left(1\,-\,\cos^2x\right)(\sin x)\left(\cos^2x\right)\,dx \;= \;\L\int\)$\displaystyle \left(\cos^2x\,-\,\cos^4x\right)(\sin x\,dx)$

Now let $\displaystyle \,u\,=\,\cos x\;\;\Rightarrow\;\;du\,=\,-\sin x\,dx$ . . . etc.

 

[calchere]

Ah, so i try to keep them the same.
Thank you very much!