- #1

karush

Gold Member

MHB

- 3,269

- 5

$\displaystyle\int_{0}^{\pi/8}\dfrac{\sec^2(2x)}{2+\tan\left({2x}\right)}$

ok not sue if this u v is correct or maybe better...

$u=\dfrac{\tan\left(2x\right)}{\sqrt{2}}\therefore du=\sqrt{2}\sec^2(2x) dx$