Answer to Question #175014 in Calculus for Phyroe

Question

∫ x tan² 3x² dx

Answer 1

$\displaystyle\int x \tan^2 (3x^2) dx$

Note that if

$d(3x^2)=(3x^2)'dx=6xdx$

then

$xdx=\dfrac{1}{6}\cdot d(3x^2)$

$\displaystyle\int x\cdot \tan^2 (3x^2) dx=$

$\displaystyle\int \tan^2 (3x^2)\cdot x dx=$

$\displaystyle\int \tan^2 (3x^2)\cdot \dfrac{1}{6}\cdot d(3x^2)=$

$\dfrac{1}{6}\cdot\displaystyle\int \tan^2 (3x^2)\cdot d(3x^2)$

Further substitute

$\tan (3x^2)=t$

Hence

$3x^2=\arctan(t)$

$d(3x^2)=d(\arctan(t))=(\arctan(t))'dt=\dfrac{1}{t^2+1}dt$

Continue solving

$\dfrac{1}{6}\cdot\displaystyle\int \tan^2 (3x^2)\cdot d(3x^2)=$

$\dfrac{1}{6}\cdot\displaystyle\int t^2 \cdot \dfrac{1}{t^2+1}dt=$

$\dfrac{1}{6}\cdot\displaystyle\int \dfrac{t^2}{t^2+1}dt=$

$\dfrac{1}{6}\cdot\displaystyle\int \dfrac{t^2+1-1}{t^2+1}dt=$

$\dfrac{1}{6}\cdot\displaystyle\int \left(\dfrac{t^2+1}{t^2+1}-\dfrac{1}{t^2+1}\right)dt=$

$\dfrac{1}{6}\cdot\displaystyle\int \left(1-\dfrac{1}{t^2+1}\right)dt=$

$\dfrac{1}{6}\cdot\displaystyle \left(t-\arctan(t)\right)+c=$ <return to variable $x$: $t=\tan (3x^2)$ >

$\dfrac{1}{6}\cdot\displaystyle \left(\tan (3x^2)-\arctan(\tan (3x^2))\right)+c=$

$\dfrac{1}{6}\cdot\displaystyle \left(\tan (3x^2)-3x^2\right)+c=$

$\dfrac{1}{6}\tan (3x^2)-\dfrac{1}{6}\cdot3x^2+c=$

$\dfrac{1}{6}\tan (3x^2)-\dfrac{x^2}{2} +c$