A Integration Problem[Indefinite Integral]

[Hysteria X]

Homework Statement

Evaluate

<br /> \int (x sin ( x^2 ) )/cos^3 ( x^2 ) dx<br />

Homework Equations

none the i know that would be helpful here

The Attempt at a Solution

I tried solving this by substitution method by substituting <br /> t = x^2 as t but i don't know how to proceed after that

 

[LCKurtz]

Homework Statement

Evaluate

<br /> \int (x sin ( x^2 ) )/cos^3 ( x^2 ) dx<br />

Homework Equations

none the i know that would be helpful here

The Attempt at a Solution

I tried solving this by substitution method by substituting <br /> t = x^2 as t but i don't know how to proceed after that

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

 

[Hysteria X]

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

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

 

[LCKurtz]

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

 

[SammyS]

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

Don't forget the constant of integration.

 

[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.

 

[Hysteria X]

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.

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?

 

[STEMucator]

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?

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?

 

[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.