Integral trigonometrica de sen, cos e cotg

[Fernandobertolaccini]

Resolver a integral :

\(\int (sen3x.cos3x-e^{5x+1}+ cotg7x)dx\)


Resp: \(\frac{1}{6}sen^23x-\frac{1}{5}e^{5x+1}+\frac{1}{7}ln(sen7x) + C\)

Muito obrigado !!

[Walter R]

Observe que \(sen3xcos3x=\frac{1}{2}(sen(3x-3x)+sen(3x+3x))=\frac{1}{2}sen6x\).

Então \(\int \frac{1}{2}sen6xdx=\frac{1}{12}\int sen6x(6dx)=-\frac{1}{12}cos6x+k_1=-\frac{1}{12}cos(2.3x)+k_1=-\frac{1}{12}\left ( 1-2sen^23x \right )+k_1=\frac{1}{6}sen^23x+k_1'\)


Temos também que \(\int -e^{5x+1}dx=-\frac{1}{5}\int e^{5x+1}5dx=-\frac{1}{5}e^{5x+1}+k_2\).

Finalmente, \(\int cotg7xdx=\frac{1}{7}\int cotg7x(7dx)=\frac{1}{7}ln|sen7x|+k_3\).