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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

$\displaystyle\int \sin(3x)\cos(5x)dx$

anyways this is how i solved it

$\displaystyle\int\frac{1}{2}\sin(3x-5x)+\frac{1}{2}\sin(3x+5x)dx$

$\displaystyle\int\frac{1}{2}\sin(-2x)+\frac{1}{2}\sin(8x)dx$

$\displaystyle\frac{1}{2}\int\sin(-2x)dx+\frac{1}{2}\int\sin(8x)dx$

$\displaystyle u=-2x$; $\displaystyle du=-2dx$; $\displaystyle dx=-\frac{1}{2}du$

$\displaystyle v=8x$; $\displaystyle dv=8dx$; $\displaystyle dx=\frac{1}{8}dv$

$\displaystyle-\frac{1}{4}\int\sin(u)du+\frac{1}{16}\int\sin(v)dv$

$\displaystyle -\frac{1}{4}(-\cos(-2x))+\frac{1}{16}(-\cos(8x))+C$

$\displaystyle \frac{1}{4}\cos(-2x)-\frac{1}{16}\cos(8x)+C$ ---- this is my answer.

the -2x part in this answer is bothering me because in my book it's not negative. please tell me why is that.