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

3 Answers
Jun 6, 2017

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

Jun 6, 2017

# 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.!

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.!