# A Integration Problem[Indefinite Integral]

1. Jan 28, 2013

### Hysteria X

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

$$\int (x sin ( x^2 ) )/cos^3 ( x^2 ) dx$$

2. Relevant equations

none the i know that would be helpful here

3. The attempt at a solution

I tried solving this by substitution method by substituting $$t = x^2$$ as t but i dont know how to proceed after that

2. Jan 28, 2013

### LCKurtz

Show us what you got, because you should be able to push that through. Alternatively, try $t=\cos(x^2)$.

3. Jan 28, 2013

### Hysteria X

Damn how could i not see that
okay i got the answer as $I= 1/4(1/cos^2(x^2)$

Can anyone confirm this :uhh:

4. Jan 28, 2013

### LCKurtz

You don't need someone else to confirm it. Differentiate your answer and see if it works.

5. Jan 28, 2013

### SammyS

Staff Emeritus
Don't forget the constant of integration.

6. Jan 29, 2013

### HallsofIvy

If you had gone with your first idea, $t= x^2$ so that dt= 2x dx and (1/2)dt= x dx, you would have got $(1/2)\int sin(t)/cos^3(t) dt$. Now, it should be clear, with that "sin(t)dt" in the numerator, you can let y= cos(t), so that dy= sin(t) dt, and the integral becomes $(1/2)\int dy/y^3= (1/2)\int y^{-3}dy$.

Essentially, you are doing the same thing in two steps rather than one.

7. Jan 29, 2013

### Hysteria X

Sir correct me if i am wrong but i think there is a mistake in the method you have done which i first stated.

if we took $t= x^2$ and dt =2x dx we cannot substitute that in the equation because $x^2$ is within the sine function and dx is outside???

8. Jan 29, 2013

### Zondrina

Wait wait wait. Listen to what you're saying here. You said t = x2 right? So everywhere x2 appears in your equation you replace it with t.

Now dt = 2x dx which implies that (1/2)dt = xdx. So anywhere you see xdx ( which it's VERY apparent to see ), you replace it with (1/2)dt.

Clear now?

9. Jan 29, 2013

### CompuChip

You replace all occurrences of x separately.

So the sine and cosine contain an x², which you substitute by t.

Then outside the trig functions, you are left with an x dx which - as you said - you substitute by (1/2) dt.