Question

`int sqrt((a+x)/(a-x))dx`?

Answer 1

x + c

Explanation:

`int sqrt((a+x)/(a-x)) dx` is to be integrated ,so lets rationalize it.

`int sqrt((a+x)/(a-x)` x `sqrt((a-x)/(a+x)) dx`

=`int sqrt( a^2-x^2)/sqrt(a^2-x^2) dx`=`int 1 dx`= x+c

Answer 2

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

Explanation:

Let, `I=intsqrt((a+x)/(a-x)) dx.`

Rationalising, we have,

`I=intsqrt((a+x)/(a-x))xx sqrt((a+x)/(a+x)) dx`

`=int(a+x)/sqrt(a^2-x^2) dx`

`=aint1/sqrt(a^2-x^2) dx -1/2int(-2x)/sqrt(a^2-x^2) dx`

`=a*arc sin(x/a)-1/2int{d/dx(a^2-x^2)*(a^2-x^2)^(-1/2)} dx`

`=a*arc sin(x/a)-1/2(a^2-x^2)^(-1/2+1)/(-1/2+1)`

`=a*arc sin(x/a)-(a^2-x^2)^(1/2)`

`rArr I=a*arc sin(x/a)-sqrt(a^2-x^2)+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.!

Answer 3

`-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.!