If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Integration by parts: ∫𝑒ˣ⋅cos(x)dx

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.E (LO)
,
FUN‑6.E.1 (EK)
Worked example of finding an indefinite integral by applying integration by parts twice, and then obtaining an equation for the desired indefinite integral. Created by Sal Khan.

## Want to join the conversation?

• Could the answer be simplified to e^x/2, due to sin x + cos x = 1?
• careful, sin x + cos x does not equal 1. sin^2(x) + cos^2(x) = 1
• Is the word integration means the same as antiderivatives ?
• Simon, you're correct, but I think your explanation is far too cumbersome to be of any help the person who may ask this kind of question.
Simply put, yes. Integrating and finding the anti-derivative mean the same thing.
• So taking derivatives of sin and cosine would end up in an infinite loop? i.e.
``Each entry is the derivative of the last.sinx , cosx, -sinx, - cosx , sinx ...  until  ∞``
• Yes, exactly.
``sin(x)cos(x)-sin(x)-cos(x)``

The interesting part about taking the derivative of `sin(x)` over and over and over is when you begin to have to solve dy^941/d^941x sin(x) = ?.

(Also, a personal milestone, this is my 800th answer! Yay!)
• Is there a way to solve this problem with the tabular method?
• I'm just learning this stuff myself, but I believe you have to be able to differentiate something down to zero in order to use the tabular method.
• Can anybody please tell me why whenever I try to change which is f(x) and g'(x), the answer is different?
• In this case the answer is the same, but in many cases one choice will make things more complicated and you will only get an answer by making a mistake!

ADDENDUM: It is also important to note that the substitution that works for the second integration depends on what substitution you did in the first integration.
• How would you integrate by parts for lets say, arc-cosine or arc-tangent?
• you would use the identity property of multiplication to make arccos into 1•arccos, then use integration by parts.
⌠arccosx dx=x•arccosx +⌠x/√(1-x²) dx (u=1-x²) = x•arccosx +(1/4)√(1-x²) + C
⌠arctanx dx=x•arctanx +⌠x/(1+x²) dx (u=1+x²) = x•arctanx + (1/2)ln|1+x²| + C
or x•arctanx + ln|√(1+x²)| + C
• Does anyone know a website with exercises of this topic?
• I am confused that why this question can't just leave it at integral by parts once, why need it twice? I mean, isn't the first time clear enough?
(1 vote)
• After the first time we apply the integration by parts method to this expression we get an answer that includes another integral. We haven't solved the problem until we have an answer that doesn't include an integral.
• Mathematica gives the answer as `(e^x * cos(x))/(ln x)`. Is this correct (it probably is) and if so, how complicated is the simplification?
• I plotted them both and this is not correct (try x=8),
f(8)=e^8*(sin(8)+cos(8))/2=1257. ....
g(8)=(e^8 * cos(8))/(ln 8)=-208. ....

The functions also behave differently at x=1, yours is very clearly 0 (due to the division by ln(0)=0, while Khans is very clearly about 1/2: (cos(1)*e^(1) / 2=0.5*2.7/2=0,7).

in mathematica, make sure to press 'esc' 'e' 'e' and again 'esc' to enter the 'e' function. For Cos use 'Cos[x]' (so capital letter, and those square brackets).