Question

an integral.

Answer 1

\[\int\limits_{ }^{ }{\rm Sin}^{-1}({\tiny~}\sqrt{x}{\tiny~~})~dx\]

Answer 2

So I am going to use an integration by parts.
I will differentiate the sine function:

\[y={\rm Sin}^{-1}({\tiny~}\sqrt{x}{\tiny~~})\]\[\sqrt{x}={\rm Sin}({\tiny~}y{\tiny~~})\]then the derivative,
\[\frac{1}{2\sqrt{x}}=y'\cos(y)\]\[\frac{1}{2\sqrt{x}}=y'\sqrt{x+1}\]\[\frac{1}{2\sqrt{x^2+x}}=y'\]

Answer 3

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

Answer 4

Call \(u=\sqrt{x}\), then \(du = \frac{dx}{2\sqrt{x}}\), so that your integral now becomes:
\[2*\int\limits_{}^{}u*asin(u)du\], now just solve by parts

Answer 5

it is an inverse sin.

Answer 6

yes asin = sin^-1

Answer 7

Not just sin.

Also, the hint I was given is to use by parts.

Answer 8

oh, I was like... sorry

Answer 9

arcsin.. lol

Answer 10

hahah

Answer 11

\[\int\limits_{ }^{ }{\rm Sin}^{-1}(\sqrt{x})~dx=x{\rm Sin}^{-1}(\sqrt{x})-2\int\limits_{ }^{ }\frac{x}{\sqrt{x^2+x}}~dx\]
should be correct, if I didn't make silly errors.

Answer 12

I will find \[-2\int\limits_{ }^{ }\frac{x}{\sqrt{x^2+x}}~dx\] and put it together.

Answer 13

what u did is correct

Answer 14

oh, tnx

Answer 15

wait, \[u=x^2+x\] doesn';t work

Answer 16

Let me see if I can find this one. give me a bit time to think

Answer 17

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

Answer 18

\[u=x^2+x\]for the first one:
\[\int\limits_{ }^{ }\frac{(1/2)}{\sqrt{u}}du-\int\limits_{ }^{ }\frac{1/2}{\sqrt{x^2+x}}~dx\]

Answer 19

i don't think you should compare, cuz i made a substituion that makes things different

Answer 20

then, the first integral is going to be:
\[\sqrt{x^2+x}+C-\int\limits_{ }^{ }\frac{1/2}{\sqrt{x^2+x}}~dx\]

Answer 21

I think it isn't going bad.

Answer 22

Then I will solve \[\frac{1/2}{\sqrt{x^2+x}}~dx\]
(I will gather the pieces later)

Answer 23

ok let's try to solve that integral

Answer 24

\[\int\limits_{ }^{ }\frac{1/2}{\sqrt{x^2+x}}~dx\]

wait, you can't use partical fractions, correct?

Answer 25

you need to solve \[\int\limits_{}^{} \frac{ x }{ \sqrt{x^2 + x} }\]

Answer 26

i think you can't in this case

Answer 27

why do I have to solve that one, and not the one I posted?

Answer 28

your expression for the result of integration by parts only needs to solve that last integral so that you get teh answer

Answer 29

alright, lets do it from here.\[\int\limits_{ }^{ }\frac{x}{\sqrt{x^2+x}}~dx\]without completing the fraction.

Answer 30

OK, let me think

Answer 31

\[\int\limits_{ }^{ }\frac{\sqrt{x}}{\sqrt{x+1}}~dx\]if that makes it any easier, but I doubt it.

Answer 32

oh wait,.

Answer 33

\[\int\limits_{ }^{ }\frac{\sqrt{x}}{\sqrt{x+1}}~dx\]
u=x+1
x=u-1
dx=du

Answer 34

\[\int\limits_{ }^{ }\frac{\sqrt{u-1}}{\sqrt{u}}~du\]

Answer 35

not much -:( but at least something

Answer 36

how do you feel about trig substitution @idku?

Answer 37

i really don't see a simple way for solving this one

Answer 38

which ?

Answer 39

trig sub for \[\int\limits_{ }^{ }\frac{x}{\sqrt{x^2+x}}~dx\]

Answer 40

complete the square inside the sqrt( ) thingy
and then apply a trig sub

Answer 41

are you familiar with trig subs?

Answer 42

\[\int\limits_{ }^{ }\frac{x}{\sqrt{(x+\frac{1}{4})^2}}~dx\]

Answer 43

x^2+x+(1/2)^2-(1/2)^2
(x+1/2)^2-1/4

Answer 44

Oh I left something out

Answer 45

\[\int\limits_{ }^{ }\frac{x}{\sqrt{(x+\frac{1}{4})^2-\frac{1}{16}}}~dx\]Am I right?

Answer 46

also here is another way to look at the integral if you prefer
\[\int\limits \sin^{-1}(\sqrt{x}) dx \\ \text{ Let } y=\sin^{-1}(\sqrt{x}) \\ \sin(y)=\sqrt{x} \\ \cos(y) dy=\frac{1}{2 \sqrt{x}} dx \\ 2 \sqrt{x} \cos(y) dy =dx \\ 2 \sin(y) \cos(y) dy =dx \\ \sin(2y) dy=dx \\ \int\limits_{}^{} y \sin(2y) dy\]

Answer 47

I don't think that is correct

Answer 48

oh then by parts for the last thing you wrote?

Answer 49

\[x^2+x=x^2+x+(\frac{1}{2})^2-(\frac{1}{2})^2=(x+\frac{1}{2})^2-(\frac{1}{2})^2 \\ x^2=x=(x+\frac{1}{2})^2-\frac{1}{4}\]

Answer 50

yes you can do integration by parts there

Answer 51

and i think it will be tons easier than this other way

Answer 52

I think I got it from here.
If anything I can use wolfram to check. ty
that was indeed easier.

Answer 53

this is also what @M4thM1nd wrote above
sorry didn't see that

Answer 54

didn't see he did it, but weither way I got the approach.

Answer 55

w should be there

Answer 56

tnx again

Answer 57

i got to go now, srry