# Integrals

• Mar 27th 2008, 12:04 AM
allenr34
Integrals
http://img168.imageshack.us/img168/7011/asdlu6.png
and
http://img174.imageshack.us/img174/2051/97318225kp3.png
Are the only ones left in my assignment, please help me if you can, i haven't got much time. Thank you
• Mar 27th 2008, 12:35 AM
TheEmptySet
I love this proceedure
We will use integration by parts twice and then solve for the integral

use $u=e^{3x} \mbox{ and }dv=\sin(x)$

$\int e^3x \sin(x)dx=-e^{3x} \cos(x)+3 \int e^{3x} \cos(x)dx$

Now we will integrate by parts again with

$u=e^{3x} \mbox{ and } dv=\cos(x)$

$\int e^3x \sin(x)dx=-e^{3x} \cos(x)+3[e^{3x} \sin(x) -3 \int e^{3x}\sin(x)dx]$

distributing and simplifying we get

$\int e^3x \sin(x)dx=-e^{3x} \cos(x)+3e^{3x} \sin(x) -9 \int e^{3x}\sin(x)dx$

now we adding $9 \int e^{3x}\sin(x)dx$ to BOTH sides of the equation.

$10\int e^3x \sin(x)dx=-e^{3x} \cos(x)+3e^{3x} \sin(x)$

Solving for the integral we get...

$\int e^{3x}\sin(x)dx=\frac{e^{3x}}{10}\left( 3\sin(x)-\cos(x)\right)$
• Mar 27th 2008, 12:41 AM
allenr34
Thank you so much, and btw, the second task, which i posted, is not the one i need to solve, i solved that myself. This is the one:
http://img258.imageshack.us/img258/4448/dasdasdcy7.png
• Mar 27th 2008, 12:55 AM
TheEmptySet
$\int \frac{\cos(3x)}{\sqrt{8+\cos^2(3x)}}dx$

Note that the denominator can be rewritten as follows

$8+\cos^2(3x)=8 + (1-1)+\cos^2(3x)=9+\cos^2(3x)-1=9-\sin^2(3x)$

so now we have the integral

$\int \frac{\cos(3x)}{\sqrt{9-\sin^2(3x)}}dx$

now using $u=\frac{1}{9}\sin(3x) \mbox{ and } du=\frac{1}{3}\cos(3x)dx$

$\int \frac{3du}{\sqrt{9-9u^2}}=\int \frac{du}{\sqrt{1-u^2}}$

You should be able to finish from here.

Good luck
• Mar 27th 2008, 01:03 AM
TheEmptySet
$u=2+\sin^3(x) \mbox{ and } du=3\sin^2(x)\cos(x)dx$

$\int \frac{\cos(x) \sin^2(x)}{\sqrt[3]{2+sin^3(x)}}dx=\frac{1}{3}\int u^{-1/3}du = \frac{1}{2}u^{2/3}$

$=\frac{1}{2} \sqrt[3]{(2+sin^3(x))^2}$

P.s if you have new questions post them in a new thread.

(Rock)