1. ## Definite integrals.

i need help with these problems please?:

(bSa is the intergration symbol with the a and b values)

1S0 36dx/(2x+1)^3

(pi)/2S0 5sin^(3/2)xcosxdx

(pi)/4S0 e^(tanx)sec^2xdx

2. Originally Posted by zdawg
1S0 36dx/(2x+1)^3
Just clarifying $\displaystyle \displaystyle\int_0^1 \frac{36}{(2x+1)^3}~dx$ ?, if so use a substitution $\displaystyle \displaystyle u= 2x+1$

3. yeah, thats what i ment, sorry.

4. Well as I pointed out $\displaystyle \displaystyle\int_0^1 \frac{36}{(2x+1)^3}~dx$

Make $\displaystyle \displaystyle u= 2x+1\implies \frac{du}{dx} = 2$

Giving $\displaystyle \displaystyle\int_0^1 \frac{2\times 18}{(2x+1)^3}~dx = \int_1^3 \frac{ 18}{(u)^3}~du= 18\int_1^3 \frac{ du}{(u)^3}$

5. my txtbk says the final answer is 8, do you know how to get there?

6. Originally Posted by zdawg
my txtbk says the final answer is 8, do you know how to get there?
I do, by completeing the hints given in post #4. Can you finish it using simple integration?

7. Originally Posted by zdawg
[snip]

(pi)/2S0 5sin^(3/2)xcosxdx

(pi)/4S0 e^(tanx)sec^2xdx
(b) Substitute $\displaystyle u = \sin (x)$.

(c) Substitute $\displaystyle u = \tan (x)$.

If you need more help, post what you can do and make it clear where you get stuck.