1. reverse subsitution (integration)

Hello everyone,
im having trouble with this question. it says find the integration of this by using reverse subsitution.
I know we take tan^3x as u and sec^2 and du...but i dont know how to continue..

Thanks.

2. Originally Posted by Oasis1993
Hello everyone,
im having trouble with this question. it says find the integration of this by using reverse subsitution.
I know we take tan^3x as u and sec^2 and du...but i dont know how to continue..

Thanks.
$\displaystyle \int \tan^3(x) \sec^2(x)\,dx$

$\displaystyle u = \tan(x) \: \: \rightarrow \: \: du = \sec^2(x)\,dx$

$\displaystyle \int \tan^3(x) \sec^2(x)\,dx = \int u^3 \, du$

Which is much easier

3. thank you!
so is the answer to this ((tan^3x)^4) / 4?

4. Almost.

$\displaystyle \frac{1}{4} \tan^{12}(x) + C$

The +C is very important and can cause you to lose the answer mark in an exame

$\displaystyle (\tan^3(x))^4 = ([\tan(x)]^3)^4 = \tan^{12}(x)$

5. thank you!
Could i also ask you another question regarding reverse subsitution?

here again.. 1-x^4 is u and 2x^3 is du?
but how does this one continue...? :S

6. Originally Posted by Oasis1993
thank you!
Could i also ask you another question regarding reverse subsitution?

here again.. 1-x^4 is u and 2x^3 is du?
but how does this one continue...? :S
$\displaystyle \int \frac{2x^3}{1-x^4} = 2 \int \frac{x^3}{1-x^4}$

There is a crafty shortcut in this example.

Set $\displaystyle u = 1-x^4$. When we differentiate with respect to x we get $\displaystyle du = 4x^3 \, dx$

$\displaystyle 2\int \frac{x^3}{u} \cdot \frac{du}{4x^3}$

This simplifies to $\displaystyle \frac{1}{2}\int \frac{du}{u}$ which is a standard integral