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# integral by substitution

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\int \frac{sin^2\theta \:cos\theta }{\sqrt{1+sin\theta }}

Jun 25, 2021

#1
+115952
+3

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

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.

Jun 25, 2021
#2
+152
+3

[quote:Melody]  $u^2=1+sin\theta$  [quote]

how about just $u=\sin(\theta)$ ?

its raised in the power of two from the thing itself

and i do not see how you would get that $1$ except when applying the common derivative

care explaining? :)

#3
+115952
+3

I wanted to get rid of the square root off the bottom.

if

$$u^2=1+sin\theta\\ \text{then denominator becomes u}$$

It was a bit of a guess but I tried it and it appeared to work.

Do you want me to explain more?

Melody  Jun 25, 2021
edited by Melody  Jun 25, 2021
edited by Melody  Jun 25, 2021
#4
+152
+3

oh i see -- that works as well; i didnt think of that LOL. did it all the way around!

#5
+115952
+3

ok, if you want me to work through it just ask but if you can do it yourself, (with mine or any other substitution) that is even better.

If you do it with a different substitution, and you get it out, please share that with us.

Melody  Jun 25, 2021
#6
+152
+3

[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

Jun 25, 2021
#7
+32832
+2

Here's my alternative substitution:

Jun 25, 2021
edited by Alan  Jun 25, 2021
#8
+115952
+1

Very nice, thanks guys.

Jun 25, 2021
#9
+26228
+3

Integral

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

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

Jun 25, 2021
edited by heureka  Jun 25, 2021
#10
+121095
+1

Very nice  work  by  everyone, here......!!!!!

CPhill  Jun 25, 2021