# Substituting definate integrals

• Aug 10th 2007, 06:54 PM
driver327
Substituting definate integrals
I have been having problems substituting and one instance of basic integration. 1 and 3 are subs and 2 is the basic integration (I believe)
1. /int (sqrt pi)_0 x(cos x)^2 the answer is 0
2. /int (x^2)(2x^3)^1/2 the answer is (2/9)(sqrt 2)(x^(9/2))+C
3. /int ((x^2)(2x^3-5)^1/2) the answer is (1/9)(2x^3-5)^3/2+C

For 2. i laid it out as x^2*(2x^3)^1/2. mult. for it to be, just 2x^3 which i know is wrong. plz help.
• Aug 10th 2007, 07:27 PM
galactus
For #1, you can use parts. $\displaystyle \int[xcos^{2}(x)]dx$

Let $\displaystyle u=x, \;\ dv=cos^{2}(x)dx, \;\ du=dx$$\displaystyle , \;\ v=\int[cos^{2}(x)]dx=\frac{xsin(x)cos(x)}{2}+\frac{x^{2}}{2}=\frac{x sin(2x)+2x^{2}}{4} For #2: \displaystyle \int[x^{2}\sqrt{2x^{3}}]dx Let \displaystyle u=2x^{3}, \;\ du=6x^{2}dx, \;\ \frac{1}{6}du=x^{2}dx For #3: \displaystyle \int[x^{2}\sqrt{2x^{3}-5}]dx Same substitutions as above, except use \displaystyle u=2x^{3}-5 • Aug 13th 2007, 01:20 PM driver327 Quote: Originally Posted by galactus For #1, you can use parts. \displaystyle \int[xcos^{2}(x)]dx Let \displaystyle u=x, \;\ dv=cos^{2}(x)dx, \;\ du=dx$$\displaystyle , \;\ v=\int[cos^{2}(x)]dx=\frac{xsin(x)cos(x)}{2}+\frac{x^{2}}{2}=\frac{x sin(2x)+2x^{2}}{4}$

For #2: $\displaystyle \int[x^{2}\sqrt{2x^{3}}]dx$

Let $\displaystyle u=2x^{3}, \;\ du=6x^{2}dx, \;\ \frac{1}{6}du=x^{2}dx$

For #3: $\displaystyle \int[x^{2}\sqrt{2x^{3}-5}]dx$

Same substitutions as above, except use $\displaystyle u=2x^{3}-5$

I get #1 and #3, for #2, can you tell me the steps to get that particular answer?
• Aug 13th 2007, 01:59 PM
Soroban
Hello, driver327!

Quote:

$\displaystyle 2)\;\; \int x^2(2x^3)^{\frac{1}{2}}dx$

The answer is: .$\displaystyle \frac{2\sqrt{2}}{9}x^{\frac{9}{2}} + C$

We have: .$\displaystyle x^2(2x^3)^{\frac{1}{2}} \;=\;x^2\cdot2^{\frac{1}{2}}\cdot x^{\frac{3}{2}} \;=\;\sqrt{2}\,x^{\frac{7}{2}}$

Now integrate: .$\displaystyle \sqrt{2}\int x^{\frac{7}{2}}\,dx$