- #1
askor
- 169
- 9
What is ##\int \tan 2x \ dx##?
What I get is
##\int \tan 2x \ dx##
##= \int \frac{\sin 2x}{\cos 2x} dx##
##= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx##
let u = sin x then ##\frac{du}{dx} = \cos x## or du = cos x dx
So
##= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx##
##= \int \frac{2u}{1 - 2u^2} du##
What I get is
##\int \tan 2x \ dx##
##= \int \frac{\sin 2x}{\cos 2x} dx##
##= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx##
let u = sin x then ##\frac{du}{dx} = \cos x## or du = cos x dx
So
##= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx##
##= \int \frac{2u}{1 - 2u^2} du##