Question

What is the indefinite integral of x*sqrt((1-x)/(1+x)) ?

Answer 1

\[\int\limits x \sqrt{\frac{1-x}{1+x}} dx ?\]

Answer 2

Yes

Answer 3

ok i think I might have an idea
i haven't fully explored it

Answer 4

put x=cos theta
dx=-sin theta d theta

Answer 5

but try multiplying the inside by (1-x)/(1-x)

Answer 6

the inside of the square root

Answer 7

then we are gong to try a trig sub

Answer 8

\[\int\limits \frac{ x(1-x)}{\sqrt{1-x^2}} dx \\ \text{ use } x= \sin(\theta)\]

Answer 9

Oh, that makes a lot of sense, I'll try that

Answer 10

I like that freckles.

Answer 11

and thanks :)

Answer 12

\[ \int\limits \cos \theta \sqrt{\frac{ 1-\cos \theta }{ 1+\cos \theta }}*-\sin \theta d \theta\]
\[=\int\limits -\sin \theta \cos \theta \sqrt{\frac{ 2\sin ^2\frac{ \theta }{ 2 } }{ 2\cos ^2\frac{ \theta }{ 2 } }}d \theta \]
\[=-\int\limits 2\sin \frac{ \theta }{ 2 }\cos \frac{ \theta }{ 2 }\frac{ \sin \frac{ \theta }{ 2 } }{ \cos \frac{ \theta }{ 2 } } \cos \theta d \theta \]
\[=-\int\limits 2 \sin ^2\frac{ \theta }{ 2 }\cos \theta d \theta \]
\[=-\int\limits \left( 1-\cos \theta \right)\cos \theta d \theta \]
\[=-\int\limits \cos \theta d \theta+\frac{ 1 }{ 2 }\int\limits 2\cos ^2 \theta d \theta \]
=?

Answer 13

Completing what surjithayer did, I would get: \[\frac{1}{2}\left(\frac{1}{2}\left(2θ+\sin \left(2θ\right)\right)\right)-\sin \left(θ\right)+C\]

Answer 14

This is what I got so far using freckles method, can someone check if I did it right?

Answer 15

your last line of your work in that pdf bothers me

Answer 16

everything else in that pdf is good til that last line

Answer 17

\[-\cos(\theta)-\frac{\theta}{2}+\frac{1}{4} \sin(2 \theta)+C \\ \text{ or } \\ -\cos(\theta)-\frac{\theta}{2}+\frac{1}{4} 2 \sin(\theta) \cos(\theta)+C \\ \text{ now recall } x=\sin(\theta) \\ \text{ so } \cos(\theta)=\sqrt{1-x^2} \\ \text{ inputtin these should give } \\ -\sqrt{1-x^2}-\frac{\arcsin(x)}{2}+\frac{1}{2}x \sqrt{1-x^2}+C\]

Answer 18

Ahhh, ok, I was having some trouble putting x = sin(theta) back into the equation, but the way you did it makes sense

Answer 19

following surjithayer's work...
\[- \int\limits \cos(\theta) d \theta+\frac{1}{2} \int\limits 2 \cos^2(\theta) d \theta \\ - \int\limits \cos(\theta) d \theta +\frac{1}{2} \int\limits (1+ \cos(2 \theta)) d \theta \\ - \sin(\theta)+\int\limits \frac{1}{2}d \theta+\frac{1}{2} \int\limits \cos(2 \theta) d \theta \\ -\sin(\theta)+\frac{1}{2} \theta+\frac{1}{4} \sin(2 \theta)+C \\ -\sin(\theta)+\frac{1}{2} \theta+\frac{1}{4} 2 \sin(\theta) \cos(\theta)+C \\ -\sin(\theta)+\frac{\theta}{2}+\frac{1}{2} \sin(\theta) \cos(\theta)+C \\ \text{ where he chose } x=\cos(\theta) \\ \text{ so you would have } \sqrt{1-x^2}=\sin(\theta) \text{ here } \\ -\sqrt{1-x^2}+\frac{\arccos(x)}{2}+\frac{1}{2} \sqrt{1-x^2} x +C\]

you should see these as the answer since arcsin(x)=-pi/2-arccos(x)

Answer 20

same answer*

Answer 21

if you aren't sure how I'm getting the other trig function in terms of x given the substitution
you can always draw a right triangle

Answer 22

Awesome, thank you so much for your help!

Answer 23

I understand your reasoning completely, it was just that I was blanking on the trig identity sin(2x) = 2sin(x)cos(x)

Answer 24

for example
\[\text{ if } x=\sin(\theta) \\ \text{ or if } \sin(\theta)=\frac{x}{1}\]
we have:

using the Pythagorean theorem
we should see we have to solve the following equation:
\[adj^2+x^2=1^2 \\ adj^2+x^2=1 \\ adj^2=1-x^2 \\ adj=\sqrt{1-x^2}\]