$$ \int sin(x).sin(3x).sin(5x)dx $$
$$ \int - \frac{sin(9x)-sin(7x)-sin(3x)+sin(x)}{4}dx $$
$$- \frac{1}{4} \int sin(9x)dx+ \frac{1}{4} \int sin(7x)+ \frac{1}{4} \int sin(3x)- \frac{1}{4} \int sin(x) $$
$u=9x , u=7x , u=3x$
$$ \int sin(9x)dx= \frac{1}{9} \int sin(u)=> \int sin(u)=-cos(u)==> \frac{-cos(u)}{9}= \frac{-cos(9x)}{9} $$
$$ \int sin(7x)dx= \frac{1}{7} \int sin(u)=> \int sin(u)=-cos(u)==> \frac{-cos(u)}{7}= \frac{-cos(7x)}{7}$$
$$ \int sin(3x)dx= \frac{1}{3} \int sin(u)=> \int sin(u)=-cos(u)==> \frac{-cos(u)}{3}= \frac{-cos(3x)}{3}$$
$$ \int sin(x)=-cosx $$
باالحاق ضرایب انها به صورت زیر میشود
$$ \frac{cos(9x)}{36}- \frac{cos(7x)}{28} - \frac{cos(3x)}{12}+ \frac{cos(x)}{4}+c $$
