Answer to Question #175014 in Calculus for Phyroe

Question #175014

∫ x tan² 3x² dx


1
Expert's answer
2021-04-15T07:52:09-0400

xtan2(3x2)dx\displaystyle\int x \tan^2 (3x^2) dx


Note that if

d(3x2)=(3x2)dx=6xdxd(3x^2)=(3x^2)'dx=6xdx

then

xdx=16d(3x2)xdx=\dfrac{1}{6}\cdot d(3x^2)


xtan2(3x2)dx=\displaystyle\int x\cdot \tan^2 (3x^2) dx=


tan2(3x2)xdx=\displaystyle\int \tan^2 (3x^2)\cdot x dx=

tan2(3x2)16d(3x2)=\displaystyle\int \tan^2 (3x^2)\cdot \dfrac{1}{6}\cdot d(3x^2)=


16tan2(3x2)d(3x2)\dfrac{1}{6}\cdot\displaystyle\int \tan^2 (3x^2)\cdot d(3x^2)


Further substitute

tan(3x2)=t\tan (3x^2)=t

Hence

3x2=arctan(t)3x^2=\arctan(t)

d(3x2)=d(arctan(t))=(arctan(t))dt=1t2+1dtd(3x^2)=d(\arctan(t))=(\arctan(t))'dt=\dfrac{1}{t^2+1}dt



Continue solving

16tan2(3x2)d(3x2)=\dfrac{1}{6}\cdot\displaystyle\int \tan^2 (3x^2)\cdot d(3x^2)=


16t21t2+1dt=\dfrac{1}{6}\cdot\displaystyle\int t^2 \cdot \dfrac{1}{t^2+1}dt=


16t2t2+1dt=\dfrac{1}{6}\cdot\displaystyle\int \dfrac{t^2}{t^2+1}dt=


16t2+11t2+1dt=\dfrac{1}{6}\cdot\displaystyle\int \dfrac{t^2+1-1}{t^2+1}dt=


16(t2+1t2+11t2+1)dt=\dfrac{1}{6}\cdot\displaystyle\int \left(\dfrac{t^2+1}{t^2+1}-\dfrac{1}{t^2+1}\right)dt=


16(11t2+1)dt=\dfrac{1}{6}\cdot\displaystyle\int \left(1-\dfrac{1}{t^2+1}\right)dt=


16(tarctan(t))+c=\dfrac{1}{6}\cdot\displaystyle \left(t-\arctan(t)\right)+c= <return to variable xx: t=tan(3x2)t=\tan (3x^2) >


16(tan(3x2)arctan(tan(3x2)))+c=\dfrac{1}{6}\cdot\displaystyle \left(\tan (3x^2)-\arctan(\tan (3x^2))\right)+c=


16(tan(3x2)3x2)+c=\dfrac{1}{6}\cdot\displaystyle \left(\tan (3x^2)-3x^2\right)+c=


16tan(3x2)163x2+c=\dfrac{1}{6}\tan (3x^2)-\dfrac{1}{6}\cdot3x^2+c=


16tan(3x2)x22+c\dfrac{1}{6}\tan (3x^2)-\dfrac{x^2}{2} +c



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