[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
