How Do You Integrate ##\int \tan 2x \ dx##?

[askor]

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

$= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx$
$= \int \frac{2u}{1 - 2u^2} du$

 

[Samy_A]

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

$= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx$
$= \int \frac{2u}{1 - 2u^2} du$

There may be easier ways to do this, but since you tried it this way, notice that $2u du= -\frac{1}{2}d(1-2u²)$

 

[SteamKing]

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

$= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx$
$= \int \frac{2u}{1 - 2u^2} du$

Rather than saying u = sin x, use u = 2x instead. Just expand tan u into $\frac{sin\, u}{cos\, u}$.
This integral is much easier to solve.

Expanding sin 2x and cos 2x in terms of sin x and cos x just makes things more complicated.

 

[HallsofIvy]

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

$= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx$
$= \int \frac{2u}{1 - 2u^2} du$

Let $v= 1- 2u^2$ so that $dv= -4u du$. Then $(-1/2)dv= 2u du$ and the integral becomes $-\frac{1}{2}\int \frac{dv}{v}$