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!