A community for students.
Here's the question you clicked on:
 0 viewing
AravindG
 2 years ago
integral (e^x(xcos x+sin x))dx=?
AravindG
 2 years ago
integral (e^x(xcos x+sin x))dx=?

This Question is Closed

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \large \int\limits e^x(xcos x+\sin x)dx\]

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1I have been stuck on this question for sometime..I am missing something somewhere.

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1@ganeshie8 , @jim_thompson5910 Can you help?

divu.mkr
 2 years ago
Best ResponseYou've already chosen the best response.0its a case in integration of by parts!!

surjithayer
 2 years ago
Best ResponseYou've already chosen the best response.0\[=\int\limits (x \cos x) e ^{x} dx+\int\limits (\sin x) e ^{x}dx\] \[=x \cos x e ^{x}\int\limits \left\{ x \left( \sin x \right)+\cos x *1 \right\}e ^{x} dx+\[\sin x e ^{x}\int\limits \cos x e ^{x}dx\]+c\] there is some mistake,try to find out.

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1I am thinking of doing it in the form e^x(f(x)+f'(x)) By supplying terms and subtracting.

surjithayer
 2 years ago
Best ResponseYou've already chosen the best response.0i have tried that also.

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1@surjithayer How did you get integral (xcos x)e^x dx=(xcos x)(e^x)

divu.mkr
 2 years ago
Best ResponseYou've already chosen the best response.0you are doing that right and its integration is e^x.f(x)+c

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1@divu.mkr can you explain?

Euler271
 2 years ago
Best ResponseYou've already chosen the best response.1everything i wrote is gone

Euler271
 2 years ago
Best ResponseYou've already chosen the best response.1it all ends up being in terms of:\[\int\limits_{}^{}e^x cosx dx= \frac{ 1 }{ 2 }e^x(sinx + cosx) + C\] and\[\int\limits_{}^{}e^x sinx dx= \frac{ 1 }{ 2 }e^x(sinx  cosx) + C\]. a rather simple integration by parts. and seperating it like surji suggested and integrating only the first term by parts one gives:\[\frac{ x }{ 2 }e^x(sinx+cosx)  \frac{ 1 }{ 2 }\int\limits_{}^{} e^x (sinx + cosx)dx + C + \int\limits_{}^{}e^x sinx dx\]

Euler271
 2 years ago
Best ResponseYou've already chosen the best response.1was my own fault. i pressed a wrong key and then backspace

Euler271
 2 years ago
Best ResponseYou've already chosen the best response.1the integral of that, by parts once, gives the first two terms of the last line. taking u = x dv = e^xcosx

Euler271
 2 years ago
Best ResponseYou've already chosen the best response.1i wouldn't mind solving it entirely if you're not convinced :)

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1Its just I want to to do this qn in form e^x((fx)+f'(x)) Format and not apply parts.

Euler271
 2 years ago
Best ResponseYou've already chosen the best response.1i'm not familiar with that method at all. only parts

Euler271
 2 years ago
Best ResponseYou've already chosen the best response.1did you want to see parts or you know how?

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1yeah I want to see parts.

Euler271
 2 years ago
Best ResponseYou've already chosen the best response.1i guess the only part is the:\[\int\limits_{}^{} e^x cosx dx\] u = e^x; du = e^x dx dv = cosx dx; v = sinx \[\int\limits_{}^{} e^x cosx dx = e^xsinx  \int\limits_{}^{} e^x sinx dx\] u = e^x; du = e^x dx dv = sinx dx; v = cosx \[\int\limits_{}^{}e^x cosx dx= e^x sinx  \left( e^x cosx + \int\limits_{}^{}e^x cosxdx \right)\] \[\int\limits_{}^{}e^x cosx dx = e^x(cosx + sinx)  \int\limits_{}^{}e^x cosx dx\]\[2\int\limits_{}^{}e^x cosx dx = e^x(sinx + cosx)\]\[\int\limits_{}^{}e^x cosx dx = \frac{ 1 }{ 2 }e^x (sinx + cosx)\]

divu.mkr
 2 years ago
Best ResponseYou've already chosen the best response.0do you have the answer..? i got one :D

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1yes final answer is 1/2e^x(xsin x+xcos xcos x)+c

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.1@Euler271 Yeah I get it now thanks!

surjithayer
 2 years ago
Best ResponseYou've already chosen the best response.0i have tried that also.

ybarrap
 2 years ago
Best ResponseYou've already chosen the best response.0Lots of integration by parts. Work on this in pieces (I've reused u and v several times but hopefully you can tell which is which by context): $$ \int e^x(x\cos x+\sin x)dx\\ \int xe^x \cos x ~dx+\int xe^x \sin x~dx\\ \text{For }\int xe^x \cos x ~dx\\ \text{Let }u=x,dv=e^x\cos x\\du=x,v=\int e^x\cos x dx\\ \text{For }\int e^x\cos x dx\\ \text{Let }u=e^x,dv=\cos x\\ du=e^x,v=\sin x\\ \text{Then }uv\int vdu=\\ e^x\sin x\int e^x\sin x dx\\ \int e^x\sin x dx=\\ \text{Let }u=e^x,dv=\sin x\\ \text{then }du=e^x,v=\cos x\\ uv\int vdu = \\ e^x\cos x+\int e^x\cos x dx\\ \text{Combining these results:}\\ \int e^x\cos x=\cfrac{e^x}{2}(\sin x+\cos x)\\ \text {So we now have }\\ \int xe^x \cos x ~dx\\ =\cfrac{xe^x}{2}(\sin x + \cos x)\int \cfrac{e^x}{2}(\sin x + \cos x)dx+\\ \qquad \cfrac{e^x}{2}(\sin x  \cos x)\\ \text{Take, }\int\frac{e^x}{2}(\sin x + \cos x)dx\\ \int\frac{e^x}{2}\sin x + \int \cfrac{e^x}{2}\cos xdx\\ =\cfrac{e^x}{4}(\sin x cos x)+\cfrac{e^x}{4}(\sin x +cos x)\\ =\cfrac{e^x}{2}\sin x\\ \text{So, }\\ \int xe^x \cos x ~dx\\ =\cfrac{xe^x}{2}(\sin x + \cos x)\cfrac{e^x}{2}\sin x+\cfrac{e^x}{2}(\sin x  \cos x)\\ =\cfrac{e^x}{2}(x\sin x + (x1)\cos x)\\ $$ Whew!! That's it!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.