int sqrt((a+x)/(a-x))dx∫√a+xa−xdx?

3 Answers
Wricheek
Jun 6, 2017

x + c

Explanation:

int sqrt((a+x)/(a-x)) dx∫√a+xa−xdx is to be integrated ,so lets rationalize it.

int sqrt((a+x)/(a-x)∫√a+xa−x x sqrt((a-x)/(a+x)) dx√a−xa+xdx

=int sqrt( a^2-x^2)/sqrt(a^2-x^2) dx∫√a2−x2√a2−x2dx=int 1 dx∫1dx= x+c

Ratnaker Mehta
Jun 6, 2017

a*arc sin(x/a)-sqrt(a^2-x^2)+C.a⋅arcsin(xa)−√a2−x2+C.

Explanation:

Let, I=intsqrt((a+x)/(a-x)) dx.I=∫√a+xa−xdx.

Rationalising, we have,

I=intsqrt((a+x)/(a-x))xx sqrt((a+x)/(a+x)) dxI=∫√a+xa−x×√a+xa+xdx

=int(a+x)/sqrt(a^2-x^2) dx=∫a+x√a2−x2dx

=aint1/sqrt(a^2-x^2) dx -1/2int(-2x)/sqrt(a^2-x^2) dx=a∫1√a2−x2dx−12∫−2x√a2−x2dx

=a*arc sin(x/a)-1/2int{d/dx(a^2-x^2)*(a^2-x^2)^(-1/2)} dx=a⋅arcsin(xa)−12∫{ddx(a2−x2)⋅(a2−x2)−12}dx

=a*arc sin(x/a)-1/2(a^2-x^2)^(-1/2+1)/(-1/2+1)=a⋅arcsin(xa)−12(a2−x2)−12+1−12+1

=a*arc sin(x/a)-(a^2-x^2)^(1/2)=a⋅arcsin(xa)−(a2−x2)12

rArr I=a*arc sin(x/a)-sqrt(a^2-x^2)+C.⇒I=a⋅arcsin(xa)−√a2−x2+C.

Note that, the later integral has been derived using the following

useful Result

Result : int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c, n!=-1.

Enjoy Maths.!

Ratnaker Mehta
Jun 8, 2017

-a*arc cos(x/a)-sqrt(a^2-x^2)+C.

Explanation:

As an Aliter, let us solve this using a Trgo. Substn.

x=acos 2y rArr dx=-2asin 2ydy.

Also, sqrt{(a+x)/(a-x)}=sqrt{(a+acos2y)/(a-acos2y)}=coty.

:. I=int (cot y)(-2asin 2y)dy.

=-2aint2cos^2ydy

=-2aint(1+cos2y)dy

=-2a(y+sin(2y)/2)

=-a(2y)-asin2y

=-a(2y)-sqrt(a^2-a^2cos^2 2y)

rArr I=-a*arc cos(x/a)-sqrt(a^2-x^2)+C.

Enjoy Maths.!

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