Integrals

http://img168.imageshack.us/img168/7011/asdlu6.png
and
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Are the only ones left in my assignment, please help me if you can, i haven't got much time. Thank you

We will use integration by parts twice and then solve for the integral

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

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

Now we will integrate by parts again with

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

$\displaystyle \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

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

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

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

Solving for the integral we get...

$\displaystyle \int e^{3x}\sin(x)dx=\frac{e^{3x}}{10}\left( 3\sin(x)-\cos(x)\right)$

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

$\displaystyle \int \frac{\cos(3x)}{\sqrt{8+\cos^2(3x)}}dx$

Note that the denominator can be rewritten as follows

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

so now we have the integral

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

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

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

You should be able to finish from here.

Good luck

$\displaystyle u=2+\sin^3(x) \mbox{ and } du=3\sin^2(x)\cos(x)dx$

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

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

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

(Rock)