integral by substitution

\(\int \frac{sin^2\theta \:cos\theta }{\sqrt{1+sin\theta }}\;d\theta\)

let   \(u^2=1+sin\theta\)

If you need more help then show us what you have managed to do on your own. 

------------------------------------------------

Ps.   I have done this almost to the end. Unless I have made some really silly mistake/s, it appears to work fine.

[quote:Melody]  If you do it with a different substitution, and you get it out, please share that with us.  [quote]

$ \int \frac{\sin ^2\left(θ\right)\cos \left(θ\right)}{\sqrt{1+\sin \left(θ\right)}}dθ $

fisrst of all lets apply the common derivative: 

we know that $ \frac{d}{dθ}\left(\sin \left(θ\right)\right) \ \ \Rightarrow  \cos \left(θ\right) $ right? 

so you have $du=\cos \left(θ\right)d\theta $ but what you need is $d theta$ so for it we get:
$dθ=\frac{1}{\cos \left(θ\right)}du $

now plugging that in in our main:
$ \int \frac{u^2\cos \left(θ\right)}{\sqrt{1+u}}\cdot \frac{1}{\cos \left(θ\right)}du $

that;s it for the right side -- for the left side you let $  u=sin(\theta ) $ ; thus:

$ \int \frac{u^2\cos \left(θ\right)}{\sqrt{1+u}}\cdot \frac{1}{\cos \left(θ\right)}du \ \ \ \Rightarrow \ \ \  \frac{u^2\cos \left(θ\right)}{\sqrt{1+u}}\cdot \frac{1}{\cos \left(θ\right)} $
 
that whole thing now gives you:
$ \frac{u^2\cos \left(θ\right)\cdot \:1}{\sqrt{1+u}\cos \left(θ\right)}  $

you can get rid of both $\cos (\theta)$ up and down, respectively $ \int \frac{u^2\cancel{\cos \left(θ\right)} \cdot \:1}{\sqrt{1+u}\cancel{\cos \left(θ\right)}} \ \ \ \Rightarrow \ \ \ \int \frac{u^2}{\sqrt{1+u}} $

lets work this part -- we have got $  \frac{d}{du}\left(1+u\right) $ yes?

that gives you
$  \frac{d}{du}\left(1\right)+\frac{d}{du}\left(u\right) $
 derative of a constant is $=0$ thus,you get $0+1$ or just $1$

now we have $ dv=1du $ but again, we are looking for $du$, so we get:  $du=1dv   $

knowing that we can set up $ \int \frac{u^2}{\sqrt{v}}\cdot \:1dv  $

to make it 'workable' we write $u$ as a difference of $v$ and $1$, since $1+u=v$

thus you end up with $\int \frac{\left(v-1\right)^2}{\sqrt{v}}dv  $

you can apply perfect square formula there, so you get $ \int \frac{v^2-2v+1}{\sqrt{v}} dv   \ \ \overset{\text{we can just write it as 3 different parts instead of one} }{=====================\Rightarrow} \ \ \ \int \frac{v^2}{\sqrt{v}}-\frac{2v}{\sqrt{v}}+\frac{1}{\sqrt{v}} dv $

now you can write $ \sqrt{v}  $ as $ v^\frac{1}{2}  $ so by doing $ v^{2-\frac{1}{2}}$ you get :
$  \int \:v^{\frac{3}{2}}-\frac{2v}{\sqrt{v}}+\frac{1}{\sqrt{v}}dv   $

you can do the same thing for $  \frac{2v}{\sqrt{v}} $ which we  get $ \frac{2v}{v^{\frac{1}{2}}} $ and again continuing we get $ 2v^{-\frac{1}{2}+1}  $ which is equal to $ 2v^{\frac{1}{2}}  $ OR JUST $ 2\sqrt{v} $

thus, you have $ \int \:v^{\frac{3}{2}}-2\sqrt{v}+\frac{1}{\sqrt{v}}dv  $

just a few last steps -- find the derative of each of the three separated parts (as we can write $\int \:v^{\frac{3}{2}}-2\sqrt{v}+\frac{1}{\sqrt{v}}dv$ \ \ \ \Rightarrow \ \ \  $\int \:v^{\frac{3}{2}}dv-\int \:2\sqrt{v}dv+\int \frac{1}{\sqrt{v}}dv $)

lets work each one of them invidually, as if id do all of them at the same time it would be a mess -- respectively:
1) $\int \:v^{\frac{3}{2}}dv \ \ \Rightarrow \ \ \  \frac{v^{\frac{3}{2}+1}}{\frac{3}{2}+1} \ \ \Rightarrow \ \ \ \frac{v^{\frac{5}{2}}}{\frac{5}{2}} \ \ \Rightarrow \ \ \ \frac{v^{\frac{5}{2}}\cdot \:2}{5}  \Rightarrow \frac{2}{5}v^{\frac{5}{2}}  $
there is nothing more you can do to that, so yeah lets continue with the second one:

2) $ \int \:2\sqrt{v}dv $ -- do not forget; always take the constant out:
$ 2\cdot \int \sqrt{v}dv \ \ \Rightarrow \ \ \  2\cdot \int \:v^{\frac{1}{2}}dv \ \ \ \Rightarrow \ \ \ 2\cdot \frac{v^{\frac{1}{2}+1}}{\frac{1}{2}+1} \ \ \ \Rightarrow \ \ \ 2\cdot\frac{v^{\frac{3}{2}}}{\frac{3}{2}}  $

i do not know if one row can handle all that so i will continue to this second row:
$  2\cdot\frac{v^{\frac{3}{2}}}{\frac{3}{2}} \ \ \Rightarrow \ \ \  2\cdot\frac{v^{\frac{3}{2}}\cdot \:2}{3} \ \ \ \Rightarrow \ \ \ \frac{4v^{\frac{3}{2}}}{3}  \ \ \ \Rightarrow \ \ \  \frac{4}{3}v^{\frac{3}{2}}   $ 

thats it done for the 2nd part -- now lets do 3rd:
3) $ \int \frac{1}{\sqrt{v}}dv \ \ \ \Rightarrow \ \ \ \int \frac{1}{v^{\frac{1}{2}}}dv \ \ \ \Rightarrow \ \ \ \int \:v^{-\frac{1}{2}}dv \ \ \ \Rightarrow \ \ \ \frac{v^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} \ \ \ \Rightarrow \ \ \ \frac{v^{-\frac{1}{2}+1}}{\frac{1}{2}} \ \ \ \Rightarrow \ \ \ \frac{v^{\frac{1}{2}}}{\frac{1}{2}} \ \ \ \Rightarrow \ \ \ \frac{v^{\frac{1}{2}}\cdot \:2}{1} \ \ \ \Rightarrow \ \ \ 2v^{\frac{1}{2}}  $

or just $2\sqrt{v}$

by putting it all together we get $\frac{2}{5}v^{\frac{5}{2}}-\frac{4}{3}v^{\frac{3}{2}}+2\sqrt{v}$

in the beginning, we said that $v=u+1,$ thus we get:
$ \frac{2}{5}(u+1)^{\frac{5}{2}}-\frac{4}{3}(u+1)^{\frac{3}{2}}+2\sqrt{u+1} $

but, we also said that $u=\sin(\theta)  $so by again substituting we get:

$ \frac{2}{5}\left(\sin \left(θ\right)+1\right)^{\frac{5}{2}}-\frac{4}{3}\left(\sin \left(θ\right)+1\right)^{\frac{3}{2}}+2\sqrt{\sin \left(θ\right)+1}  $

never forget to add the constant after integrating:
$ \frac{2}{5}\left(\sin \left(θ\right)+1\right)^{\frac{5}{2}}-\frac{4}{3}\left(\sin \left(θ\right)+1\right)^{\frac{3}{2}}+2\sqrt{\sin \left(θ\right)+1} +C $

SO our answer is $ \boxed{\frac{2}{5}\left(\sin \left(θ\right)+1\right)^{\frac{5}{2}}-\frac{4}{3}\left(\sin \left(θ\right)+1\right)^{\frac{3}{2}}+2\sqrt{\sin \left(θ\right)+1} +C} $

:D
 

Here's my alternative substitution:

Very nice, thanks guys.    

Integral

\(\int \dfrac{\sin^2(\theta) \cos(\theta) }{\sqrt{1+\sin(\theta) }}~d\theta\)

\(\begin{array}{|l|c|c|} \hline & \text{substitute} \\ \hline \int \dfrac{\sin^2(\theta) \cos(\theta) }{\sqrt{1+\sin(\theta)}}~d\theta & u = \sin(\theta)& du = \cos(\theta)~d\theta \\ & & d\theta =\dfrac{du}{\cos(\theta)} \\ =\int \dfrac{u^2 \cos(\theta) }{\sqrt{1+u}}*\dfrac{du}{\cos(\theta)} \\\\ =\int \dfrac{u^2 }{\sqrt{1+u}}~du & v=1+u & dv = du \\ & u=v-1& du = dv \\\\ =\int \dfrac{(v-1)^2 }{\sqrt{v}}~dv & w=\sqrt{v} & dw = \dfrac{1}{2\sqrt{v}}~dv \\ &v=w^2 & dv = 2\sqrt{v}~dw \\\\ =\int \dfrac{(w^2-1)^2 }{\sqrt{v}}~2\sqrt{v}~dw \\ =2\int(w^2-1)^2 ~dw \\\\ =2\int ( w^4-2w^2+1) ~dw \\\\ =2\left( \dfrac{w^5}{5}-\dfrac{2w^3}{3}+w\right) ~+c \\\\ =\dfrac{2w^5}{5}-\dfrac{4w^3}{3}+2w ~+c & w & \\ & =\sqrt{v} \\ & =\sqrt{1+u} \\ & =\sqrt{1+\sin(\theta)} \\\\ =\dfrac{2}{5}\Big(\sqrt{1+\sin(\theta)}\Big)^5 \\ -\dfrac{4}{3}\Big(\sqrt{1+\sin(\theta)}\Big)^3 \\ +2\sqrt{1+\sin(\theta)} ~+c \\\\ \hline \end{array}\)