# Integral Help!

1. Aug 8, 2011

### stargazer843

∫1+sinx/(cosx)² dx

I made u = cos x.

du = -sinxdx

-du = sinxdx

so:

∫1+sinxdx/(cosx)² = ∫1-du/u² = ∫ (1/u²) * (1-du)

This is where I got stuck. the 1-du is throwing me off. distributing would get me nowhere and I don't know how to get rid of the 1.

2. Aug 8, 2011

### Dr. Seafood

We have a https://www.physicsforums.com/forumdisplay.php?f=152"!

Is the integrand (1 + sin x)/cos2x ? Or 1 + sin x/cos2x?

In the second case, for the second addend note that if u = cos x, -du = sin x dx:

∫ 1 + sin x/cos2x dx = x + ∫ -du/u2 = x + 1/u + C = x + 1/cos x + C

In the first case, write 1/cos2x + sin x/cos2x and integrate the second term as we did previously.
Also, 1/cos2x = sec2x -- this is the derivative of which function?

Last edited by a moderator: Apr 26, 2017
3. Aug 8, 2011

### stargazer843

first case, and ill remember that next time, although this isn't homework

4. Aug 8, 2011

### stargazer843

Once I get it to ∫1-du/u²

If I separate it:

[∫(1/u²)]-[∫(du/u²)]

[∫(1/u²)du]-[∫(1/u²)du]

This = 0

I can only separate it at the beginning?

5. Aug 8, 2011

### rock.freak667

Instead of substitution where you would get something like dx = 1-du, which is not helpful, why don't we rewrite the integral as

$$\int \left( \frac{1}{cos^2x} + \frac{sinx}{cos^2x} \right) dx$$

Now do you know what 1/cosx is the same as and what sinx/cosx is the same as?

When you get that, rewrite it the integral again but replace 1/cosx with the equivalent and do the same with sinx/cosx.

Post what you get.

6. Aug 8, 2011

### Staff: Mentor

This is incorrect, in part because it is meaningless.
Each integral needs the differential factor. In what you have above, ∫(1/u²) is missing du.

7. Aug 8, 2011

### stargazer843

ah ok, now I understand.

Thank you for all the help guys! :D