Question #6b7c7

4 Answers
Aug 30, 2017

#int sqrt ((a+x)/(a-x))dx=asin^-1 (x/a)-sqrt (a^2-x^2)+c#
Where #c# is the constant of integration.

Explanation:

#int sqrt ((a+x)/(a-x))dx#
Rationalizing by multiplying numerator and denominator both by #(a+x)# we get
#int(a+x)/sqrt (a^2-x^2)dx#
#=inta/sqrt (a^2-x^2)dx +intx/sqrt (a^2-x^2)dx#
#=asin^-1(x/a) -1/2 int ((-2x))/sqrt (a^2-x^2)dx#
#=asin^-1 (x/a)-1/2×2sqrt (a^2-x^2)+c#
#=asin^-1 (x/a)-sqrt (a^2-x^2)+c#
Where #c# is the constant of integration.

Aug 30, 2017

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

Explanation:

Let, #I=intsqrt{(a+x)/(a-x)}dx.#

Rationalising, we get, #I=intsqrt{(a+x)/(a-x)xx(a+x)/(a+x)}dx,#

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

#=aint1/sqrt(a^2-x^2)dx+intx/sqrt(a^2-x^2)dx,#

#=a*arc sin(x/a)-1/2int(-2x)/sqrt(a^2-x^2)dx,#

#=a*arc sin(x/a)-1/2int(a^2-x^2)^(-1/2)*d/dx(a^2-x^2)dx,#

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

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

Note that, the latter integral follows from,

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

#=ln|f(x)|+c, if, n=-1.#

Enjoy Maths.!

Aug 30, 2017

#-sqrt((a-x)(a+x))-2asin^-1(sqrt((a-x)/(2a)))+C#

Explanation:

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

Let #u=sqrt(a-x)#. This implies that #u^2=a-x# so #2udu=-dx# (we'll use the opposite version of this). It also implies that #x=a-u^2#. The integral then becomes:

#I=intsqrt(a+a-u^2)/u(-2udu)=-2intsqrt(2a-u^2)du#

Now let #u=sqrt(2a)sintheta#. This implies that #du=sqrt(2a)costhetad theta#.

Importantly, note that #2a-u^2=2a-2asin^2theta=2a(1-sin^2theta)=2acos^2theta#.

#I=-2intsqrt(2acos^2theta)(sqrt(2a)costhetad theta)=-2(2a)intcos^2thetad theta#

Use #cos^2theta=1/2(1+cos2theta)#:

#I=-2aint(1+cos2theta)d theta=-2a(theta+1/2sin2theta)#

Use #sin2theta=2sinthetacostheta#:

#I=-2a(theta+sinthetacostheta)#

Recall that #sintheta=u/sqrt(2a)#. Draw the corresponding right triangle where the side opposite #theta# is #u# and the hypotenuse is #sqrt(2a)#. Through the Pythagorean theorem, the side adjacent to #theta# is #sqrt(2a-u^2)#. Thus, #costheta=sqrt(2a-u^2)/sqrt(2a)#.

Also, note that #theta=sin^-1(u/sqrt(2a))#.

#I=-2a(sin^-1(u/sqrt(2a))+u/sqrt(2a)sqrt(2a-u^2)/sqrt(2a))#

Rearranging:

#I=-usqrt(2a-u^2)-2asin^-1(u/sqrt(2a))#

Recall that #u=sqrt(a-x)#.

#I=-sqrt(a-x)sqrt(2a-(a-x))-2asin^-1(sqrt(a-x)/sqrt(2a))#

#color(white)I=color(blue)(-sqrt((a-x)(a+x))-2asin^-1(sqrt((a-x)/(2a)))+C#

Aug 30, 2017

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

Explanation:

Here is an another way to solve the Problem.

Let, #I=intsqrt{(a+x)/(a-x)}dx.#

Knowing that, #(1+cos2y)/(1-cos2y)=cot^2y=cos^2y/sin^2y,# we use

the substn. #x=acos2y,#

# rArr dx=a(-sin2y)(2)dy=-4asinycosydy.#

#:. I=intsqrt{(a+acos2y)/(a-acos2y)}*(-4asinycosy)dy,#

#=-4aintcosy/cancelsiny*cancelsinycosydy,#

#=-2aint2cos^2ydy=-2aint(1+cos2y)dy,#

#=-2a{y+sin(2y)/2},#

#=-a(2y)+asin2y,#

# :. I =-a(2y)+asqrt(1-cos^2(2y)).#

Since, #x=acos2y, 2y=arc cos(x/a),# we have,

# I=-a*arc cos(x/a)+asqrt{1-(x/a)^2},#

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

Enjoy Maths!