Answer
$-\dfrac{1}{6} \ln |\cos (2x^3)| +C$
Work Step by Step
Our aim is to solve the integral $ \int x^2 \tan (2x^3) \ dx$
Let us consider that $a =2x^3$ and $\dfrac{da}{dx}=6x^2 \implies dx=\dfrac{1}{6x^2} \ da$
Now, $ \int x^2 \tan (2x^3) \ dx = \int x^2 \tan a \times (\dfrac{1}{6x^2}) \ da$
or, $= \dfrac{1}{6} \int \tan a \ da+C$
or, $=-\dfrac{1}{6} \ln |\cos a| +C$
or, $=-\dfrac{1}{6} \ln |\cos (2x^3)| +C$
