Question

Помогите пожалуйста Интеграл!!!!!!!!!!!!!!!!!!

Answer 1

Ответ:

$a)\ \ \displaystyle \int \frac{dx}{1+cos6x}=\\\\=\Big[\ t=tg\, 3x\ ,\ cos6x=\frac{1-t^2}{1+t^2}\ ,\ 3x=arctg\, t\ ,\ dx=\dfrac{dt}{3(1+t^2)}\ \Big]=\\\\\\=\int \frac{dt}{3(1+t^2)(1+\frac{1-t^2}{1+t^2})}=\frac{1}{3}\int \frac{dt}{1+t^2+1-t^2}=\frac{1}{6}\int dt=\frac{1}{6}\cdot t+C=\\\\\\=\frac{1}{6}\cdot tg\, 3x+C$

$b)\ \ \displaystyle \int \frac{x\, dx}{sin^2x}\Big[\ u=x\ ,\ du=dx\ ,\ dv=\frac{dx}{sin^2x}\ ,\ v=-ctgx\ \Big]=uv-\int v\, du=\\\\\\=-x\cdot ctgx+\int ctgx\, dx=-x\cdot ctgx+\int \frac{cosx\, dx}{sinx}=-x\cdot ctgx+\int \frac{d(sinx)}{sinx}=\\\\\\=-x\cdot ctgx+ln\Big|sinx\Big|+C$