Integration of a trig function

[chwala]

Homework Statement

$\int\frac {tan x}{1+ cos^2x}dx$

Relevant Equations

integration

This is my first attempt ...

 

[chwala]

my second attempt...

 

[chwala]

My third attempt...bingo, I got it!

 

[Charles Link]

Your method works, but I did it with $\tan{x}=\sin{x}/\cos{x}$, and $\sin{x} \, dx=-d \cos{x}$.
With $u=\cos{x}$, the problem then becomes an exercise in partial fractions, and it also gets the same answer as that of the above post.

 

[SammyS]

my other approach...

This works nicely, but you have an error in differentiating.

The derivative of $\sec x$ is $\tan x \cdot \sec x$,

so that $du=2\tan x \cdot \sec^2 x \ dx$.

The rest of the work will follow nicely from that.

 

[chwala]

Yeah I noticed that...thanks slight error, I left out $dx$

 

[SammyS]

Yeah I noticed that...thanks slight error, I left out $dx$

You also missed squaring the secant function.

 

[chwala]

You also missed squaring the secant function.

I think you are looking at the wrong page, check my post number $3$

 

[SammyS]

I think you are looking at the wrong page, check my post number $3$

No. My replies are referring to Post # 2.

You have the following:

But you have an error in du.

It should be $du=2\tan x \cdot \sec^2 x \ dx$ .

Thus your integral becomes $\displaystyle \frac 1 2 \int \frac{du}{u}$ .

Etc.

 

[chwala]

No. My replies are referring to Post # 2.

You have the following:
View attachment 275901
But you have an error in du.

It should be $du=2\tan x \cdot \sec^2 x \ dx$ .

Thus your integral becomes $\displaystyle \frac 1 2 \int \frac{du}{u}$ .

Etc.

Noted cheers...