Question

INTEGRAL. I am practicing by making up problems. Don't always solve them, but in most the cases i do.

Answer 1

\[\int\limits_{ }^{ }\sin^{-1}\sqrt{x}~dx\]

Answer 2

So first I differentiate the inverse sine.

Answer 3

the function,\[\sin(y)=\sqrt{x}\]
the derivative,
\[y'\times \cos(y)=\frac{1}{2\sqrt{x}}\]

Answer 4

\[y'=\frac{1}{2\sqrt{x}\cos(y)}\]

Answer 5

then if sin(y)=x/1, then the adjacent side is sqrt{1-x^2}
So,
\[y'=\frac{1}{2\sqrt{x-x^3}}\]

Answer 6

So now,
u=sin^-1(sqrt x)
dv=1

Answer 7

\[\int\limits_{ }^{ }\sin^{-1}(\sqrt{x})~dx=x\sin^{-1}x - \int\limits_{ }^{ }\frac{x}{2\sqrt{x-x^3}}~dx\]

Answer 8

\[\int\limits_{ }^{ }\sin^{-1}(\sqrt{x})~dx=x\sin^{-1}x - \int\limits_{ }^{ }\frac{\sqrt{x}}{2\sqrt{1-x^2}}~dx\]

Answer 9

@jim_thompson5910

Answer 10

i was wondering if trig sub works

Answer 11

oh if I sub x=cos(theta) i get it simplfied (or maybe that is even worse)

Answer 12

I'm stuck honestly, but I'm still thinking it over

Answer 13

I mean sin theta

Answer 14

\[\int\limits\limits_{ }^{ }\frac{\sqrt{x}}{2\sqrt{1-x^2}}~dx\]\[x=\sin\theta\]\[dx=\cos\theta~d\theta\]

Answer 15

\[\int\limits\limits_{ }^{ }\frac{\sqrt{\sin\theta}}{2\sqrt{1-\sin^2\theta}}\times \cos\theta~d\theta\]

Answer 16

\[(1/2)\int\limits\limits_{ }^{ }\sqrt{\sin\theta}~d\theta\]

Answer 17

but is this simpler to integrate?

Answer 18

wolfram is getting this

http://www.wolframalpha.com/input/?i=integral(arcsin(sqrt(x)))&t=crmtb01

but honestly I'm not sure how

Answer 19

maybe it's using u-sub to say u = sqrt(x), du = dx/(2*sqrt(x)) ---> du = dx/(2u) ---> du = 2u*du

but I'm not sure how to run with that

Answer 20

from where i am now, it is deadly.
by parts won't help, will just add more theta's to the trig functions.

i did an incorrect trig sub i guess.

Answer 21

wait, what if I said x=sin^2theta from there?

Answer 22

i mean intead of just sin theta

Answer 23

oops I meant to say dx = 2u*du

but still not sure what to do with all that

Answer 24

i think i might know it...

Answer 25

\[\int\limits_{ }^{ }\frac{\sqrt{x}}{2\sqrt{1-x^2}}~dx\]\[u=\sqrt{1-x^2}\]
sqrt{1-x^2}=u
1-x^2=u^2
-x^2=(u^2-1)
no i fail to solve for sqrt(x) in terms of u. I get i.

Answer 26

I will retry, maybe i can substitute u=sqrt{x} from the very beginning. tnx for trying

Answer 27

\[\int\limits_{ }^{ }\frac{x}{2\sqrt{x}\sqrt{1-x^2}}~dx\]\[u=1-x^2\]\[x=\sqrt[{\LARGE 4}]{1-u}\]\[-\frac{1}{2}du=x~dx\]\[-\int\limits_{ }^{ }\frac{1}{\sqrt[4]{1-u}\sqrt{u}}~du\]\[-\int\limits_{ }^{ }\frac{1}{\sqrt[4]{u^2-u^3}}~du\]

Answer 28

\[\int\arcsin(\sqrt x)\,dx\]
Substituting \(u=\sqrt x\) gives \(u^2=x\), and \(dx=2u\,du\).
\[\int\arcsin(\sqrt x)\,dx\stackrel{x\to u^2}{=}2\int u\arcsin u\,du\]
Integrating by parts, you can set
\[\begin{matrix}f=\arcsin u&&&dg=u\,du\\
df=\frac{du}{\sqrt{1-u^2}}&&&g=\frac{1}{2}u^2\end{matrix}\]
\[\int u\arcsin u\,du=\frac{1}{2}u^2\arcsin u-\frac{1}{2}\color{red}{\int\frac{u^2}{\sqrt{1-u^2}}\,du}\]
For the remaining integral, you can make the substitution \(u=\sin t\), then \(du=\cos t\,dt\).
\[\color{red}{\int\frac{u^2}{\sqrt{1-u^2}}\,du}\stackrel{u\to\sin t}{=}\int\frac{\sin^2t\cos t}{\sqrt{1-\sin^2t}}\,dt=\int\sin^2t\,dt=\frac{1}{2}\int(1-\cos2t)\,dt\]

Answer 29

@idku it looks like you tried IBP from the get-go. Setting that up like so:
\[\begin{matrix}
f=\arcsin(\sqrt x)&&&dg=dx\\
df=\frac{dx}{2\sqrt{x-x^2}}&&&g=x
\end{matrix}\]
and so
\[\int \arcsin(\sqrt x)\,dx=x\arcsin(\sqrt x)-\frac{1}{4}\color{red}{\int\frac{x}{\sqrt{x-x^2}}\,dx}\]
For this integral, complete the square in the radical and make a trigo sub.
\[x-x^2=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\]
Let \(x-\dfrac{1}{2}=\dfrac{1}{2}\sin p\), giving \(dx=\dfrac{1}{2}\cos p\,dp\).
\[\begin{align*}\color{red}{\int\frac{x}{\sqrt{x-x^2}}\,dx}&\stackrel{x-\frac{1}{2}\to\frac{1}{2}\sin p}{=}\frac{1}{2}\int\frac{\left(\dfrac{1}{2}\sin p+\dfrac{1}{2}\right)\cos p}{\sqrt{\dfrac{1}{4}-\dfrac{1}{4}\sin^2p}}\,dp\\\\

&=\frac{1}{2}\int\left(\sin p+1\right)\,dp
\end{align*}\]