Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Trying to use the integration by parts trick with an extra term

the question is as follows.

$\int^{2x}_{1} y^{-1} e^{xy} \cos (y) dy$ can you solve this with integration by parts trick? it looks odd.. you obviously don't wanna integrate the y term which means u wanna derivate it then integrate in the next step with the other piece to get the stuff in the middle to be your integral but i can't seem to do it correctly.


A more clear explanation of what hes doing below is ( which i may of fudged up )

let F(x)=$\int^{2x}_{1} y^{-1} e^{xy} \cos (y) dy$ then derivative with respect to x is $F^{'}(x)$= (by chain rule) $2*(y^{-1} e^{xy} \cos (y)) + I(x)$ evaluated at $2x$ because the derivative of 1=0 $2*(2x^{-1} e^{2x^{2}} \cos (2x))$ =$(x^{-1} e^{2x^{2}} \cos (2x))$ + I(x) now I(x) is just the derivative of the function with respect x holding y constant so $\int^{2x}_{1}y*y^{-1} e^{xy} \cos (y)$ the y comes form the derivative of $e^{xy}$

now we have

$\int^{2x}_{1} e^{xy} \cos (y)dy$ by integration by parts integrate $\cos(y)$ derivate $e^{xy}$ we get $e^{xy} \sin (y)$ - $\int^{2x}_{1} xe^{xy} \sin (y)dy$ again by parts we get $e^{xy} \sin (y)$- $(- xe^{xy} \cos (y)dy)$ - $(x^{2} \int^{2x}_{1} e^{xy} \cos (y)dy)$ the second integral is the same as the first so I(x)= $e^{xy} \sin (y)$- $(- xe^{xy} \cos (y)dy)$ - $(x^{2} I(x))$

thus I(x) =$(e^{xy} \sin (y)+ xe^{xy} \cos (y))/(1+x^{2}) |^{2x}_{1}$

thus $F^{'}(x)= (x^{-1} e^{2x^{2}} \cos (2x)) + (e^{xy} \sin (y)+ xe^{xy} \cos (y))/(1+x^{2})|^{2x}_{1}$

Thus $F(x)= \int [ (x^{-1} e^{2x^{2}} \cos (2x)) + (e^{xy} \sin (y)+ xe^{xy} \cos (y))/(1+x^{2})|^{2x}_{1})] dx$

share|cite|improve this question
I think integration by parts deserves to be called a theorem. – 1015 Mar 22 '13 at 4:42
haha its not the theorem i was really inquiring about but the I trick that is written below – Faust7 Mar 22 '13 at 4:45
up vote 2 down vote accepted

Let $I(x)=\int^{2x}_{1} y^{-1} e^{xy} \cos (y) dy$. We have $I(1)=0, I'(x)=2y^{-1} e^{xy} \cos (y)+\int^{2x}_{1} e^{xy} \cos (y) dy$ The indefinite integral yields to two integrations by parts, differentiating the $\cos y$ and then the $\sin y$ that results, returning the same integral with a factor $-x^2$ so we have $\int^{2x}_{1} e^{xy} \cos (y) dy=\frac{e^{xy}(x \cos y + sin y)}{1+x^2}|_1^{2x}$ Now integrate with respect to $x$

share|cite|improve this answer
i actually understand that but there no way i can integrate that with respect to x – Faust7 Mar 22 '13 at 4:51
Did you think by any chance that the bounds of integration of $I(0)$ were equal when $x=1$? According to WA, $I(1)$ is small, but nonzero. I think you want to do the trick with $I(1/2)$. – 1015 Mar 22 '13 at 4:56
he took the derivative of the integral with respect to x making one of the bound 0 * ( stuff) and the other bound 2*(stuff) then used the a simple trick to integrate in y now wants to integrate in x then evaluate the function ( i think) – Faust7 Mar 22 '13 at 4:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.