I'm having difficulty with solving this:
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{cos\:t}{\sqrt{3+sin\:t}}\: dt$
I'd greatly appreciate any help!
Make the substitution $\displaystyle \displaystyle u = 3 + \sin{t} \implies du = \cos{t}\,dt$, and also $\displaystyle \displaystyle u(0) = 3$ and $\displaystyle \displaystyle u\left(\frac{\pi}{2}\right) = 4$ and the integral becomes
$\displaystyle \displaystyle \int_3^4{\frac{1}{\sqrt{u}}\,du} = \int_3^4{u^{-\frac{1}{2}}\,du}$.
Can you go from here?
Like Prove it said the substitution $\displaystyle u=3+\sin(t) \Rightarrow du=\cos(t)dt$ is very useful here.
Because you have a definite integral you can change the integration limits by using the given substitution.
The original integration limits are in function of $\displaystyle t$ so the new one has to be in function of $\displaystyle u$ so:
$\displaystyle t=\frac{\pi}{2} \Rightarrow u=3+\sin\left(\frac{\pi}{2}\right)=3+1=4$
$\displaystyle t=0 \Rightarrow u=3+\sin(0)=3$
So the new integration limits (in function of $\displaystyle u$) are 3 and 4.
To solve the integral use the rule:
$\displaystyle \int_{a}^{b} x^{n} = \left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}$ and $\displaystyle n\neq -1$
Note:
Changing the integration limits isn't necessary, you can also hold the original integration limits, but afterwards you have to do the back-substitution then.
You are making a substitution $\displaystyle \displaystyle u(t) = 3 + \sin{t}$, so the notation $\displaystyle \displaystyle u(0)$ means $\displaystyle \displaystyle u$ evaluated at $\displaystyle \displaystyle t = 0$, and similarly $\displaystyle \displaystyle u\left(\frac{\pi}{2}\right)$ means $\displaystyle \displaystyle u$ evaluated at $\displaystyle \displaystyle t = \frac{\pi}{2}$. As Siron said, by changing the terminals to $\displaystyle \displaystyle u$ values, you don't need to convert your integral back to a function of $\displaystyle \displaystyle t$ before substituting your terminals.