# Find the integral of sinx/cos^3x dx

1. Sep 14, 2009

### tjbateh

1. The problem statement, all variables and given/known data

Find the integral of sinx/cos^3x dx

2. Relevant equations

3. The attempt at a solution

How would I approach such a problem?

Last edited by a moderator: Jul 9, 2014
2. Sep 14, 2009

### snipez90

Re: Integration

If that's a cos(x) raised to the 3rd power in the denominator, then your integrand is just tan(x)sec^2(x) which is easy since the derivative of tan(x) is sec^2(x).

3. Sep 14, 2009

### tjbateh

Re: Integration

Wait so if I substitute U for cos(x)^3, then du=tan(x)sec^2(x)?

4. Sep 14, 2009

### Gregg

Re: Integration

No, he's saying that the integrand can be rewritten as tanxsec^2 x which can be easily integrated.

5. Sep 14, 2009

### tjbateh

Re: Integration

the integrand as in the whole problem? So sin(x)/cos(x)^3=tan(x)sec(x)^2??

6. Sep 14, 2009

### slider142

Re: Integration

The integrand of an integral is the expression between the summa $$\int$$ and the differential dx. The equation you wrote above is correct for the integrand, which you can now integrate easily.

7. Sep 14, 2009

### Gregg

Re: Integration

Yes. If

$$y=\sec^2(x)$$

Then:

$$\frac{dy}{dx} = \cdots$$

8. Sep 14, 2009

### tjbateh

Re: Integration

tan(x)sec(x)?? But i don't understand, what happens to the sin(x) in the numerator..It seems like were just talking about the denominator.

9. Sep 14, 2009

### slider142

Re: Integration

Do you remember the common definition of tan(x) = sin(x)/cos(x)?

10. Sep 14, 2009

### Gregg

Re: Integration

$$y=sec^2(x) \Rightarrow y=\frac{1}{cos(x)}\frac{1}{cos(x)}$$

Use the product rule to differentiate that, you will see you have your derivative that is the [almost] the same as the integrand. Hence you have the answer.

11. Sep 14, 2009

### tjbateh

Re: Integration

wow, it finally makes sense! Thank you everyone!

12. Sep 15, 2009

### njama

Re: Integration

Here is much simpler approach:

$$\int{\frac{sinx}{cos^3x} dx}$$

u=cos(x)

du=-sin(x)dx

dx=-du/sin(x)

$$\int{\frac{sin(x)}{u^3}*\frac{-du}{sin(x)}}=$$

$$=-\int \frac{du}{u^3}$$

13. Sep 15, 2009

### tjbateh

Re: Integration

This was more of the approach we learned in class. Would you then use the LN function?

14. Sep 15, 2009

### Bohrok

Re: Integration

You use $\ln$ if the integrand were 1/u where the denominator has a power of one, but for any other power, use the power rule for integrals.