Question

How to solve this indefinite integral? ​

Answer 1

$\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx$

Substitute $y=2x^2$, so that $\mathrm dy=4x\,\mathrm dx$:

$\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx=\frac34\int\frac{4x}{\cos^2(2x^2)}\,\mathrm dx=\frac34\int\frac{\mathrm dy}{\cos^2y}$

Then

$\dfrac1{\cos^2y}=\sec^2y=\dfrac{\mathrm d}{\mathrm dy}[\tan y]$

so that the integral wrt $y$ comes out to be

$\displaystyle\frac34\int\sec^2y\,\mathrm dy=\frac34\tan y+C$

Replace $y$ to solve for the integral wrt $x$:

$\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx=\boxed{\frac34\tan(2x^2)+C}$