1. ## Integral

To practice integration I am doing Princeton University integration exams and this question came up...I used a method on it which I am not sure is valid...if it isn't please explain why not, and if they're is an easier method please do tell

$\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{4+\sin^2(x)}d x$

Let $sin(x)=u$
THen $du=cos(x)$

so we get $\int\frac{1}{4-u^2}du=\int\frac{1}{(2+u)(2-u)}du$

Partial fractions reveals $\frac{1}{(2-u)(2+u)}=\frac{1}{4(2+u)}+\frac{1}{4(2-u)}$

integrating we get $\int\frac{1}{4(2+u)}+\frac{1}{4(2-u)}du=\frac{\ln|2+u|}{4}+\frac{\ln|2-u|}{4}$

Now back substituting we get

$\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{4-\sin^2(x)}dx=\bigg[\frac{\ln|2+\sin(x)|}{4}+\frac{\ln|2-\sin(x)|}{4}\bigg]\bigg|_0^{\frac{\pi}{2}}=\frac{1}{4}\ln(3)$

So is that right?

2. Originally Posted by Mathstud28
Partial fractions reveals $\frac{1}{(2-u)(2+u)}=\frac{1}{4(2+u)}-\frac{1}{4(2-u)}$
Hello,

I think it's

$\frac{1}{(2-u)(2+u)}=\frac{1}{4(2+u)}{\color{red}+}\frac{1}{4( 2-u)}$

3. Originally Posted by Moo
Hello,

I think it's

$\frac{1}{(2-u)(2+u)}=\frac{1}{4(2+u)}{\color{red}+}\frac{1}{4( 2-u)}$
Yeah sorry that is what I had on my paper...I dont know why I switched it

4. Originally Posted by Mathstud28
To practice integration I am doing Princeton University integration exams and this question came up...I used a method on it which I am not sure is valid...if it isn't please explain why not, and if they're is an easier method please do tell

$\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{4+\sin^2(x)}d x$

Let $sin(x)=u$
THen $du=cos(x)$

so we get $\int\frac{1}{4-u^2}du=\int\frac{1}{(2+u)(2-u)}du$

Partial fractions reveals $\frac{1}{(2-u)(2+u)}=\frac{1}{4(2+u)}+\frac{1}{4(2-u)}$

integrating we get $\int\frac{1}{4(2+u)}+\frac{1}{4(2-u)}du=\frac{\ln|2+u|}{4}{\color{red}-}\frac{\ln|2-u|}{4}$

Now back substituting we get

$\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{4-\sin^2(x)}dx=\bigg[\frac{\ln|2+\sin(x)|}{4}{\color{red}-}\frac{\ln|2-\sin(x)|}{4}\bigg]\bigg|_0^{\frac{\pi}{2}}=\frac{1}{4}\ln(3)$

So is that right?
So after that, there remains problems (in red)
But the result is ok if it was really a - sign

5. Originally Posted by Mathstud28
To practice integration I am doing Princeton University integration exams and this question came up...I used a method on it which I am not sure is valid...if it isn't please explain why not, and if they're is an easier method please do tell

$\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{4+\sin^2(x)}d x$

Let $sin(x)=u$
THen $du=cos(x)$

so we get $\int\frac{1}{4-u^2}du=\int\frac{1}{(2+u)(2-u)}du$
I'm afraid you changed the plus for the minus sign there, but the idea was ok

$\int_0^{\tfrac{\pi }
{2}} {\frac{{\cos \left( x \right)}}
{{4 + \sin ^2 \left( x \right)}}dx} = \tfrac{1}
{2} \cdot \int_0^{\tfrac{\pi }
{2}} {\frac{{\tfrac{1}
{2} \cdot \cos \left( x \right)}}
{{1 + \left( {\tfrac{{\sin \left( x \right)}}
{2}} \right)^2 }}dx} = \tfrac{1}
{2} \cdot \int_0^{\tfrac{1}
{2}} {\frac{{du}}
{{1 + u^2 }}} = \tfrac{1}
{2} \cdot \arctan \left( {\tfrac{1}
{2}} \right)
$

6. back-substituting is not an efficient method of solving these problems. You should also change the limits of the integral as well.

so the substitution $u = \sin x$ will change $\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{4+\sin^2(x)}d x
$

to $\int_{0}^{1}\frac{1}{4-u^2}du$

as when $x = \frac{\pi}{2}$ $u = \sin \frac{\pi}{2} = 1$

and when $x = 0$ $u = \sin 0 = 0$.

__________________________________________________ ______

Here is another question that examiners/universities love to ask on this topic.

Could you explain the following error in my working.

I wish to solve the integral $I = \int_{-1}^{1}{\frac{1}{1+x^2}}dx$

Using the substitution $t = \frac{1}{x}$

$dt = - \frac{1}{x^2} dx$
$\Rightarrow dt = - t^2 dx$
$dx = - \frac{1}{t^2} dt$

the limits stay the same because $t = \pm 1$ at $x = \pm 1$

so the integral becomes

$\int_{-1}^{1}{\frac{1}{1 + \frac{1}{t^2} dt} \times - \frac{1}{t^2} dt }$

$\Rightarrow - \int_{-1}^{1}{\frac{1}{1 + t^2}dt }$

so $I = - I$

therefore $I = 0$

however $\frac{1}{1+x^2} > 0$ so the integral cannot be equal to zero.

So where did I go wrong?

P.S. the question is for Mathstud28 no other takers please.

Bobak

7. Originally Posted by Mathstud28
To practice integration I am doing Princeton University integration exams and this question came up..
The integral is easy, it's an obvious arctangent. Now, why am I sayin' this, 'cause you can "see" it. First, take $\arctan \left( \frac{\sin x}{2} \right)$ and contemplate its derivative, which is $\frac{\dfrac{\cos x}{2}}{1+\dfrac{\sin ^{2}x}{4}}=\frac{2\cos x}{4+\sin ^{2}x},$ and the conclusion follows quickly.

-----

I suggest you to find nice integrals, like hard ones. (Of course, when sayin' "hard" I mean "it looks hard but with a simple trick we can kill it.")

8. Originally Posted by bobak
back-substituting is not an efficient method of solving these problems. You should also change the limits of the integral as well.

so the substitution $u = \sin x$ will change $\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{4+\sin^2(x)}d x
$

to $\int_{0}^{1}\frac{1}{4-u^2}du$

as when $x = \frac{\pi}{2}$ $u = \sin \frac{\pi}{2} = 1$

and when $x = 0$ $u = \sin 0 = 0$.

__________________________________________________ ______

Here is another question that examiners/universities love to ask on this topic.

Could you explain the following error in my working.

I wish to solve the integral $I = \int_{-1}^{1}{\frac{1}{1+x^2}}dx$

Using the substitution $t = \frac{1}{x}$

$dt = - \frac{1}{x^2} dx$
$\Rightarrow dt = - t^2 dx$
$dx = - \frac{1}{t^2} dt$

the limits stay the same because $t = \pm 1$ at $x = \pm 1$

so the integral becomes

$\int_{-1}^{1}{\frac{1}{1 + \frac{1}{t^2} dt} \times - \frac{1}{t^2} dt }$

$\Rightarrow - \int_{-1}^{1}{\frac{1}{1 + t^2}dt }$

so $I = - I$

therefore $I = 0$

however $\frac{1}{1+x^2} > 0$ so the integral cannot be equal to zero.

So where did I go wrong?

P.S. the question is for Mathstud28 no other takers please.

Bobak
Would it be because you changed to an improper integral but did not treat it as such? There is a diguised infinite discontinuity at t=0

9. Originally Posted by Krizalid
The integral is easy, it's an obvious arctangent. Now, why am I sayin' this, 'cause you can "see" it. First, take $\arctan \left( \frac{\sin x}{2} \right)$ and contemplate its derivative, which is $\frac{\dfrac{\cos x}{2}}{1+\dfrac{\sin ^{2}x}{4}}=\frac{2\cos x}{4+\sin ^{2}x},$ and the conclusion follows quickly.

-----

I suggest you to find nice integrals, like hard ones. (Of course, when sayin' "hard" I mean "it looks hard but with a simple trick we can kill it.")
Do you have any examples of these?

10. I'm currently posting the "Today's calculation of integral # $x.$" I'm selecting problems which sometimes apparent to be hard or easy but the idea is to find nice solutions.

11. Originally Posted by Mathstud28
Would it be because you changed to an improper integral but did not treat it as such?
Not sure what you mean here.

There is a diguised infinite discontinuity at t=0
everything is transparent, could you expand on the significant of his discontinuity ?

Bobak

Do you have any examples of these?

$\int \frac{1}{a^2+x^2} dx$ is a standard form, among with many others, Krizalid is calling the integral easy because if you spot the standard form (and are good with the chain rule) there is actually nothing to do, but i assume the test was intended for student that may not be familiar with it.

Contact me directly if you want some resources.

Bobak

12. Originally Posted by bobak
Not sure what you mean here.

everything is transparent, could you expand on the significant of his discontinuity ?

Bobak

$\int \frac{1}{a^2+x^2} dx$ is a standard form, among with many others, Krizalid is calling the integral easy because if you spot the standard form (and are good with the chain rule) there is actually nothing to do, but i assume the test was intended for student that may not be familiar with it.

Contact me directly if you want some resources.

Bobak
Your two integrals are equal accept at x=0 and therefore they are not the same thing..this is because although you elminated the $\frac{-1}{t^2}$ by cancelling it doesnt remove the discontinuity

13. Originally Posted by Mathstud28
Your two integrals are equal accept at x=0 and therefore they are not the same thing..this is because although you elminated the $\frac{-1}{t^2}$ by cancelling it doesnt remove the discontinuity
Originally Posted by Bobak
everything is transparent, could you expand on the significant of his discontinuity
@Bobak:
Is substituting $t = \frac1{x}$ allowed when between the limits we have x=0?

14. Originally Posted by Isomorphism
@Bobak:
Is substituting $t = \frac1{x}$ allowed when between the limits we have x=0?
I talked to Bobak later on a PM and he said that I was correct...and you are saying the same thing..the substitution makes an improper integral at 0...therefore it is not the same integral