How do you find the indefinite integral of #int x/(sqrt(9-x^2))#?

Answer 1

#intx/sqrt(9-x^2)dx=-sqrt(9-x^2)+C#

#I=intx/sqrt(9-x^2)dx#
Let #u=9-x^2#. Differentiating this shows that #du=-2xdx#. Luckily our numerator is this only off by a factor of #-2#.
#I=-1/2int(-2x)/sqrt(9-x^2)dx=-1/2int1/sqrtudu=-1/2intu^(-1/2)du#
Using #intu^ndu=u^(n+1)/(n+1)+C#, this becomes:
#I=-1/2(u^(1/2)/(1/2))=-sqrtu=-sqrt(9-x^2)+C#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#x sin^{-1}(x/3)-sin^{-1}(x/3)+c#.
By parts. Set #u(x)=x#, #{dv}/{dx}=1/sqrt(9-x^2)#

Then #{du}/{dx}=1# and #v=sin^{-1}(x/3)#. The integration by parts formula is: #int u(x)v(x) dx = u(x)v(x)-int v(x){du}/{dx}dx#
(or #int udv = uv-int vdu# if your prefer)
so that #v(x)=sin^{-1}(x/3)# giving the integral shown in the answer.

Notes:

I assume that either you can quote the standard integral #int 1/sqrt(a^2-x^2)dx=sin^{-1} (x/a)# (or #cos^{-1}(x/a)#, it makes little difference). If not, you have to remember that you can differentiate all the inverse trig functions e.g. #y=sin^{-1}x# by getting the #x# on to the left first: #x=sin y#, then differentiate with respect to #y# not #x#, getting #dx/{dy}=cos y = sqrt(1-x^2)#, then reciprocate both sides #dy/{dx}=1/(sqrt(1-x^2)#. Then the integration is reverse of differentiation.
You have to guess correctly which part to make #u(x)# and which to make #v(x)#. A simple rule of thumb here is that that the #x# on the top is an irritation and so if you differentiate it will turn into 1 and go away. If you guess wrong the #x# will turn into an #x^2# which is bad news. Similarly if the #x# had been and #x^2# you would have needed two rounds of integration by parts.
I gloss over certain issues about "principal value" and the choice of #sin^-1x# or #cos^-1x#.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To find the indefinite integral of (\int \frac{x}{\sqrt{9-x^2}} , dx), use the substitution method. Let (u = 9 - x^2), then (du = -2x , dx).

Substitute (u) and (du) into the integral:

[\int \frac{x}{\sqrt{9-x^2}} , dx = -\frac{1}{2} \int \frac{1}{\sqrt{u}} , du]

Now, integrate (\frac{1}{\sqrt{u}}):

[\int \frac{1}{\sqrt{u}} , du = 2\sqrt{u} + C]

Substitute back (u = 9 - x^2):

[= 2\sqrt{9 - x^2} + C]

So, the indefinite integral of (\int \frac{x}{\sqrt{9-x^2}} , dx) is (2\sqrt{9 - x^2} + C).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7