Answer
$-2 \ln |\cos (x^2)| +C$
Work Step by Step
Our aim is to solve the integral $ \int (4x) \tan (x^2) \ dx$
Let us consider that $a =x^2$ and $\dfrac{da}{dx}=2x \implies dx=\dfrac{1}{2x} \ da$
Now, $ \int (4x) \tan (x^2) \ dx = \int (4x) \tan a \times (\dfrac{1}{2x}) \ da$
or, $= 2 \int \tan a \ da+C$
or, $=-2 \ln |\cos a| +C$
or, $=-2 \ln |\cos (x^2)| +C$
