Integration using substitution involving Trig (1 Viewer)

[hobbsy4]

Hey

I was just wondering what u would do if you got a question where you have to use the substitution such as x = sint and the thing you're subbing it into looks the same as the the integral of inverse sinx without the -1 on the top of the question.

WOuld you try to use the integration of inverse trig of sub in sint for x???

It is an assignment so I dont want the answer, just suggestions on how to get there.

Thanks.

 

[Slidey]

Um.

The integral of arcsin(x) can be found by noting that since (uv)'=u'v+uv' (product rule), then Integral((uv)')=Integral(u'v)+Integral(uv'), or Integral(u'v)=uv-Integral(uv')
In the case of arcsin(x), u'=1, v=arcsin(x), thus v'=1/sqrt(1-x^2), u=x
So the integral is equal to x.arcsin(x)-Int(x/sqrt(1-x^2)
For this final integral use the substitution t=1-x^2, dt=-2x dx, then the answer is:
x.arcsin(x)+sqrt(1-x^2)

Does that help?

Or did you mean the deriative of arcsin(x)?
Well, y=arcsin(x), x=sin(y)
dx/dy=cos(y)
dy/dx=1/cos(y)=1/cos(arcsin(x))
Draw up a little triangle to find that arcsin(x)=arccos(sqrt(1-x^2)), thus
dy/dx=1/sqrt(1-x^2)

Also, note that the derivative of arcsin(x) is the same as the derivative of arccos(x), but of opposite sign. That is: derivative of arccos(x)=-1/sqrt(1-x^2)

If this does not help, I suggest you be a little clearer.

 

[Slidey]

Here is a question where you could use the substitution x=sint:
Integral of x/sqrt(1-x^2) dx
let x=sint, then dx=cost dt
Thus the integral becomes: Int sint.cost/sqrt(1-sin^2(t)) dt
= Int sint dt since 1-sin2(t)=cos^2(t)
= -cos(t) + C
= -cos(arcsin(x)) + C
= -sqrt(1-x^2) + C since arcsin(x)=arccos(sqrt(1-x^2))