# Evaluating an Indefinite Integral using Substitution

1. Sep 4, 2015

### Staff: Mentor

1. The problem statement, all variables and given/known data
Evaluate the Integral:
∫sin2x dx/(1+cos2x)

2. Relevant equations

3. The attempt at a solution

I first broke the numerator up:
∫2sinxcosx dx /(1+cos2x)
2∫sinxcosx dx /(1+cos2x)

Then I let u = cosx so that du = -sinx dx
-2∫u du/(1+u2)

And now I'm stuck. I thought about turning 1+u2 into tan-1u + c, but I still have another u to deal with in the numerator.

2. Sep 4, 2015

### SammyS

Staff Emeritus
Try a further substitution.

What is the derivative of 1 + u2 ?

3. Sep 4, 2015

### Staff: Mentor

Okay. So if I let y = 1+u2, then dy = 2u du, or dy/2 = u du

That gives me -2/2∫y-1 dy
-∫y-1 dy

Integrating:
-ln|y| + c

Since y = 1+u2 that becomes:
-ln|1+u2|+ c

And since u = cosx:
-ln |1 + cos2x| + c

4. Sep 4, 2015

### ahmed markhoos

hint: differentiation of $ln(x) = \frac{1}{x}$

5. Sep 4, 2015

### SammyS

Staff Emeritus
Right.

So looking back at your two substitutions, letting u = 1 + cos2(x) would have worked very nicely.

6. Sep 4, 2015

### epenguin

Er, before you do any calculus can't you very significantly simplify the thing you are asked to integrate?

7. Sep 4, 2015

### Staff: Mentor

How so? I'm not getting anything that looks useful.

u = 1 + cos2x, du = -2sinx dx
That still leaves cosx in the numerator.

I'm open to suggestions.

8. Sep 4, 2015

### SammyS

Staff Emeritus

9. Sep 4, 2015

### Staff: Mentor

Oh. Apparently it's -2sinxcosx. Yeah that would work nicely.

It's been a while since I did any differentiation and I appear to have forgotten some important details about how to use the chain rule.

10. Sep 4, 2015

### epenguin

Not only?

11. Sep 4, 2015

### Staff: Mentor

Huh?

12. Sep 4, 2015

### epenguin

OK my late night mistake..

13. Sep 4, 2015

### Staff: Mentor

No worries!

Anyways, thanks for the help all! Sammy, I think I'll have to spend a little time going of differentiation rules again...

14. Sep 4, 2015

### SammyS

Staff Emeritus
Sure. Practice helps.

It was interesting that you got the part of the chain rule that most people miss and missed the part they get.

15. Sep 5, 2015

### epenguin

Conclusion?

16. Sep 5, 2015

### SammyS

Staff Emeritus

This was his final result.
After that I suggested a single substitution to do the job, which he indicates he now adequately understands.

17. Sep 5, 2015

### epenguin

Ah yes, I had not read carefully enough, I just read 'substitution' and thought oof, that is a bit heavy. In fact in #1 he was only missing that the numerator is (apart from numerical factor) is the derivative of the denominator. I had just mentally written
$$2 \int \frac{cos x sin x}{1 + cos^2 x} dx$$
$$. = - 2 \int \frac{cos x }{1 + cos^2 x} d (cos x)$$
$$. =-. \int \frac{d (cos^2 x)}{1 + cos^2 x}.$$
$$. = - \int \frac{d (1 + cos^2 x)}{1 + cos^2 x}.$$
$$. = - ln (1 + cos^2 x) + C$$
which you can also write
$$. = - ln K (1 + cos^2 x)$$

which is just the some thing as Drakkith, just slightly differenti setting out.