# Where am i doing it wrong(integration)?

Determine the value of integral from -pie to pie [2x(1+sin(x)]/[1+cos(x)*cos(x)]

sorry i can't provide it's answer because i am not able to tex tags i tried a lot but every time i am getting error. that doesn't print things as it should. see this(next line) where am i wrong
$$\frac{2x(1+sinx)}{1+cos^2x}[\tex]. did you see it's effect.???? My steps of solving question are. (1) change x by -x(property) (2)put cos(x)=t upper and lower limmit will became equal so it's answer should zero but it's answer is not zero. It's answer is pie square. i have reached to pie square. How. After first step the integration can also be written as limit 0 to pie(do also twice of it). After that i solve it ans reached to answer pie square. Afterall my question is why it' answer came zero if i put cos(x)=t in first case. where can i found the pdf or htm file containing the way of using tex or any other such things. Last edited: ## Answers and Replies Related Calculus and Beyond Homework Help News on Phys.org phyzguy Science Advisor The second tex marker should have a forward slash (/tex), not a back slash(\tex). I just took what you typed and changed the back slash. Then it works: [tex]\frac{2x(1+sinx)}{1+cos^2x}$$.

The second tex marker should have a forward slash (/tex), not a back slash(\tex). I just took what you typed and changed the back slash. Then it works:

$$\frac{2x(1+sinx)}{1+cos^2x}$$.
https://www.physicsforums.com/showthread.php?t=386951 [Broken] I see all this from that pdf.

After all leave this tex itex issue and try to answer.

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Mark44
Mentor
https://www.physicsforums.com/showthread.php?t=386951 [Broken] I see all this from that pdf.
Near the top of the PDF there is an error in with the closing tex tag that is shown. I have reported this error, so it should be changed soon.
After all leave this tex itex issue and try to answer.
$$\int_{-\pi}^{\pi}\frac{2x(1 + sin(x))~dx}{1 + cos^2(x)}$$

By the way, the name of this Greek letter is "pi" not "pie."

Regarding the integral, I wouldn't use that substitution. Instead, I would rewrite the denominator, replacing cos2(x) by 1 - sin2(x). That would make the integrand
$$\frac{2x(1 + sin(x))}{-sin^2(x)}$$

At this point you could split the integral into two integrals:
$$-2\int x~csc^2(x)dx - 2\int x csc(x)dx$$

Integration by parts might work on these, but I haven't done the work, so don't know for sure.

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Near the top of the PDF there is an error in with the closing tex tag that is shown. I have reported this error, so it should be changed soon.

$$\int_{-\pi}^{\pi}\frac{2x(1 + sin(x))~dx}{1 + cos^2(x)}$$

By the way, the name of this Greek letter is "pi" not "pie."

Regarding the integral, I wouldn't use that substitution. Instead, I would rewrite the denominator, replacing cos2(x) by 1 - sin2(x). That would make the integrand
$$\frac{2x(1 + sin(x))}{-sin^2(x)}$$

At this point you could split the integral into two integrals:
$$-2\int x~csc^2(x)dx - 2\int x csc(x)dx$$

Integration by parts might work on these, but I haven't done the work, so don't know for sure.
but you forget to answer the point i raise.
I have also solved this question. But my question is that, that if you substitute cos(x) by t then upper and lower limits became equal in that case it's answer should zero(upper and lower limit are same). but it is not so.
thanks for reporting error.

Ray Vickson
Homework Helper
Dearly Missed
but you forget to answer the point i raise.
I have also solved this question. But my question is that, that if you substitute cos(x) by t then upper and lower limits became equal in that case it's answer should zero(upper and lower limit are same). but it is not so.
thanks for reporting error.
By the way: in tex you should use "\sin", etc., not "sin", etc. Here is the difference: $\sin$ (using "\sin") vs. $sin$ (using "sin"). This also applies to cos, tan, sec, csc, cotan, ln, log, exp, cosh, sinh, tanh and some others.

RGV

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dextercioby
Homework Helper
I haven't done the calculations, but the antiderivative of the integrand may not be expressible in terms of elementary functions. I might be wrong, though, so the OP is urged to try if the substitution $t=\tan \frac{x}{2}$ might solve

$$\int {x}\frac{1+\sin x}{2-\sin^2{x}} dx$$

Mark44
Mentor
If t = cos(x), then x = cos-1(t), provided that x $\in$ the interval [0, $\pi$]. To change the limits of integration, they need to be in this restricted domain, and -$\pi$ isn't in it. That's the best explanation I can come up with.

Mark44
Mentor
By the way: in tex you should use "\sin", etc., not "sin", etc. Here is the difference: $\sin$ (using "\sin") vs. $sin$ (using "sin"). This also applies to cos, tan, sec, csc, cotan, ln, log, exp, cosh, sinh, tanh and some others.
IMO, that's a matter of style. I don't see any advantage in using \sin etc. in LaTeX expressions over not using it.

Also, I try to use as little LaTeX as I can get away with. Pages with large amounts of LaTeX take an inordinate amount of time to render in my browser.

Ray Vickson
Homework Helper
Dearly Missed
IMO, that's a matter of style. I don't see any advantage in using \sin etc. in LaTeX expressions over not using it.

Also, I try to use as little LaTeX as I can get away with. Pages with large amounts of LaTeX take an inordinate amount of time to render in my browser.
Well, using \sin instead of sin looks much better when typeset in a document; it may not matter so much in a homework forum. However, pointing out the distinction to the OP cannot be harmful and might even be helpful. Certainly, when submitting papers to journals the publisher would likely insist upon correct usage.

RGV

If t = cos(x), then x = cos-1(t), provided that x $\in$ the interval [0, $\pi$]. To change the limits of integration, they need to be in this restricted domain, and -$\pi$ isn't in it. That's the best explanation I can come up with.
I like this explanation.

Your explanation arises another question that is
why 0 to $\pi$ why not -$\pi$ to 0 is taken as range of cos-1 function.
When i have read inverse trigonometric equation chapter it was written that, we take it from 0 to pi to make it a function. else it will not fit in definition of function.
There it was for just to make a function but that point is affecting integration's limit so i think i need to know more deeply.

Mark44
Mentor
They (I don't know who) had to restrict the domain of the cos-1 function, and the interval [0, $\pi$] was a reasonable choice. Any interval on which the function was one-to-one would have worked.