Question

how about
∫(x2)dx/√(1−(x−1)^2)
hint:
x−1=sinθ

Answer 1

\[x ^{3}/3\left| \cos \theta \right|\] ???

Answer 2

\[x= \sin \theta +1\]
\[dx=\cos \theta d\theta\]

\[∫\frac{(sin \theta +1)^2 dx}{\sqrt{1-sin^2 \theta} }=∫\frac{(sin \theta +1)^2 cos\theta d\theta}{cos^2 \theta }\]\[=∫\frac{(sin \theta +1)^2 d\theta}{cos \theta }\] then use integration by parts

Answer 3

is it \[(\sin \theta +1)\] need to expand??

Answer 4

\[*(\sin \theta +1)^2\]

Answer 5

yes I would because some of the variable will cancel out that way

Answer 6

I got the answer \[-sin \theta -2ln(cos(\theta))+c\]

Answer 7

Hey! I sort of have a question is that
\[\int\limits_{}^{}\frac{x^2}{\sqrt{1-(x-1)^2}} dx\]

Answer 8

\[x-1=\sin(\theta) => dx=\cos(\theta) d \theta\]
\[x-1=\sin(\theta) => x=\sin(\theta)+1 =>x^2=(\sin(\theta)+1)^2\]
\[\int\limits_{}^{}\frac{(\sin(\theta)+1)^2}{\sqrt{1-(\sin(\theta))^2}} \cos(\theta) d \theta\]
\[\int\limits_{}^{}\frac{(\sin(\theta)+1)^2}{\sqrt{\cos^2(\theta)}} \cos(\theta) d \theta\]
I'm going to assume cos(theta)>0
so we have
\[\int\limits_{}^{}(\sin(\theta)+1)^2 d \theta=\int\limits_{}^{}(\sin^2(\theta)+2 \sin(\theta)+1) d \theta\]

Answer 9

\[\int\limits_{}^{}\frac{1}{2}(1-\cos(2 \theta)) d \theta+2 \int\limits_{}^{}\sin(\theta) d \theta +\int\limits_{}^{}1 d \theta\]
\[\frac{1}{2}(\theta-\frac{1}{2}\sin(2 \theta))+2 (-\cos(\theta))+\theta+C\]

Answer 10

Let me clean this up a bit
After that we should write this in terms of x(since that is the way we started)

Answer 11

\[\frac{1}{2} \theta-\frac{1}{4}\sin(2 \theta)-2\cos(\theta)+\theta+C\]
\[\frac{1}{2} \theta-\frac{1}{4} \cdot 2 \sin(\theta) \cos(\theta) -2\cos(\theta)+\theta+C\]
\[\frac{1}{2} \theta-\frac{1}{2} \sin(\theta) \cos(\theta)-2 \cos(\theta) +\theta+C\]

Answer 12

Ok remember our substitution from way above
\[x-1=\sin(\theta)\]
We should draw a right triangle base on this.
Remember sin(of an angle)=opp side of that angle/hyp
So we are given
\[\sin(\theta)=\frac{x-1}{1} (=\frac{opp}{hyp})\]

We can find the other side in terms of x by using the Pythagorean Thm!

Answer 13

\[adj=\sqrt{1^2-(x-1)^2}=\sqrt{1-(x-1)^2}\]

Answer 14

So we have
\[\cos(\theta)=\frac{\sqrt{1-(x-1)^2}}{1}=\sqrt{1-(x-1)^2}=(\frac{adj}{hyp})\]

Answer 15

Back to our answer!
\[\frac{1}{2} \theta-\frac{1}{2} \sin(\theta) \cos(\theta)-2 \cos(\theta) +\theta+C \]
=
\[\frac{1}{2} \sin^{-1}(x-1)-\frac{1}{2}(x-1)\sqrt{1-(x-1)^2}-2 \sqrt{1-(x-1)^2}+\sin^{-1}(x-1)+C\]
\[=\frac{3}{2} \sin^{-1}(x-1)-\frac{1}{2}(x-1)\sqrt{1-(x-1)^2}-2 \sqrt{1-(x-1)^2}+C\]
We could probably do a little more to simplify like multipliy the second term out and then combine more like terms. I will leave that to you.

Answer 16

I believe Zed mad a mistake above. He didn't do the sqrt(cos^2(theta)).

Answer 17

made* lol

Answer 18

tq myininaya ;)

Answer 19

well done, very good job myinaya lol...happy new year to all

Answer 20

Good spot myininaya, I see what I missed.

Answer 21

Thanks :)