Question

Question #6b7c7

Answer 1

`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.

Answer 2

`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.!

Answer 3

`-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`

Answer 4

`-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!