# Integration

• Sep 15th 2008, 12:48 PM
skabani
Integration
im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

1. S p^5 ln p dp

i got ln p(1/6p^6) - 37/100 p^12 + C, but i think that is wrong

2. S 1 (x^2 + 1) e^-x dx
0

i got 1+ 3e^-1 + C

3. S b =sqrt(3) arctan (1/x) dx
a = 1

4. S b = (1) (r^3)/sqrt (4+r^2)) dr
a = 0
• Sep 15th 2008, 01:03 PM
ThePerfectHacker
Quote:

Originally Posted by skabani
im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

1. S p^5 ln p dp

$\displaystyle \int x^5 \ln x dx = \int e^{6\ln x} \ln x (\ln x)' dx$
Let $\displaystyle t=\ln x$.
• Sep 15th 2008, 01:13 PM
icemanfan
I think the answer for question 2 should be $\displaystyle -1 - 2e^{-1}$.
• Sep 15th 2008, 01:17 PM
skabani
did u go through integration through seperation twice? or did you use a different method/
• Sep 15th 2008, 01:43 PM
icemanfan
Quote:

Originally Posted by skabani
did u go through integration through seperation twice? or did you use a different method/

First I broke the integral apart: $\displaystyle \int _0 ^1 x^2e^{-x} + e^{-x} dx = \int _0 ^1 x^2e^{-x} dx + \int _0 ^1 e^{-x} dx$

Then I used integration by parts twice on the first integral to yield $\displaystyle -x^2e^{-x} - 2xe^{-x} -2 \int _0 ^1 e^{-x} dx$.

Adding the second integral gives $\displaystyle -x^2e^{-x} - 2xe^{-x} - \int _0 ^1 e^{-x} dx = -x^2e^{-x} - 2xe^{-x} + e^{-x}$ (evaulated from 0 to 1).
• Sep 15th 2008, 02:04 PM
skabani
thanks.

do you happen to know what to do for number 4, because i am completely stuck there...
• Sep 15th 2008, 02:10 PM
icemanfan
Try using a trig substitution. I solved the problem using $\displaystyle r = 2 \tan \theta$.
• Sep 15th 2008, 02:17 PM
Chris L T521
Quote:

Originally Posted by skabani
im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

...
[snip]

4. S b = (1) (r^3)/sqrt (4+r^2)) dr
a = 0

Make the trig substitution $\displaystyle r=2\tan\vartheta\implies \,dr=2\sec^2\vartheta\,d\vartheta$

Thus, the integral becomes $\displaystyle \int_0^{\tan^{-1}\left(\frac{1}{2}\right)}\frac{16\tan^3\vartheta \cdot\sec^2\vartheta\,d\vartheta}{4\sec\vartheta}= 4\int_0^{\tan^{-1}\left(\frac{1}{2}\right)}\tan^3\vartheta\sec\var theta\,d\vartheta$

(Hint: to integrate this, break off a factor of $\displaystyle \tan\vartheta$ and then apply the identity $\displaystyle \tan^2\vartheta=\sec^2\vartheta-1$; then apply the substitution $\displaystyle z=\sec\vartheta$)

Try to take it from here :D

--Chris
• Sep 17th 2008, 09:20 PM
skabani
Quote:

Originally Posted by ThePerfectHacker
$\displaystyle \int x^5 \ln x dx = \int e^{6\ln x} \ln x (\ln x)' dx$
Let $\displaystyle t=\ln x$.

im not sure how to do this still, i keep getting different anwers every time I try it by integration by parts..can you help me out?