How do you integrate by substitution #int x/(sqrt(1-x^2) dx#?

2 Answers
Roy E.
Jan 27, 2017

You don't. By inspection, the answer is #(-1/2)sqrt(1-x^2)+c#

Explanation:

It's of the form #(f'(x))/sqrt(f(x))#. In other words, the thing on the top is (nearly) the derivative of the thing under the square root. When you differentiate #sqrt(f(x)# you get #(1/2)(f'(x))/sqrt(f(x))# (chain rule).

sente
Jan 27, 2017

#intx/sqrt(1-x^2)dx=-sqrt(1-x^2)+C#

Explanation:

By substitution:

Let #u = 1-x^2#. Then #du = -2xdx#. Substituting, we get

#intx/sqrt(1-x^2)dx = -1/2int1/sqrt(1-x^2)(-2x)dx#

#=-1/2int1/sqrt(u)du#

#=-1/2intu^(-1/2)du#

#=-1/2((u^(-1/2+1))/(-1/2+1))+C#

#=-1/2((u^(1/2))/(1/2))+C#

#=-sqrt(u)+C#

#=-sqrt(1-x^2)+C#

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