Answer to Question #136206 in Calculus for qwerty

Question

\int \:\frac{1+sin\left(mx\right)}{cos^2\left(mx\right)}dx

Answer 1

$\int \:\frac{1+sin\left(mx\right)}{cos^2\left(mx\right)}dx$

replacement

$u = mx; \frac{du}{dx}=m;$ $dx = \frac{1}{m}du;$

$\frac{1}{m}\int\frac{sin(u)+1}{cos^2(u)}du =\frac{1}{m} \int\frac{sin(u)}{cos^2(u)}du+$ $\frac{1}{m}\int\frac{1}{cos^2(u)}du$ =  $=\frac{1}{m} \int\frac{sin(u)}{cos^2(u)}du+\frac{1}{m}tan(u)$

$\frac{1}{m}\int\frac{sin(u)}{cos^2(u)}du$ ; replacement $v=cosu;\frac{dv}{du}=-sin(u);du=-\frac{1}{sin(u)}dv$

$\frac{1}{m}\int\frac{sin(u)}{cos^2(u)}du =-\frac{1}{m}\int\frac{1}{v^2}dv=-\frac{1}{mv}$ =$-\frac{1}{mcos(u)}$

$\frac{1}{m}\int\frac{sin(u)+1}{cos^2(u)}du =\frac{1}{m}tan(u)-- \frac{1}{mcos(u)}$

reverse replacement u = mx

$\int \:\frac{1+sin\left(mx\right)}{cos^2\left(mx\right)}dx =$ $\dfrac{\tan\left(mx\right)}{m}+\dfrac{1}{m\cos\left(mx\right)} +C$

Answer: $\dfrac{\tan\left(mx\right)}{m}+\dfrac{1}{m\cos\left(mx\right)} +C$