Question

How I integrate this?

`int_0^(1/2)dx/(4-5sqrt(1-x^2)`

Answer 1

Rationalize the denominator then use lots of partial fraction decompositions.

Explanation:

Let

`I=int_0^(1/2)dx/(4-5sqrt(1-x^2))`

Rationalize:

`I=int_0^(1/2)dx/(4-5sqrt(1-x^2))*((4+5sqrt(1-x^2))/(4+5sqrt(1-x^2)))`

Simplify:

`I=int_0^(1/2)(4+5sqrt(1-x^2))/(25x^2-9)dx`

Integration is distributive:

`I=4int_0^(1/2)1/(25x^2-9)dx+5int_0^(1/2)sqrt(1-x^2)/(25x^2-9)dx`

Rearrange:

`I=-4int_0^(1/2)1/((3+5x)(3-5x))dx+5int_0^(1/2)(1-x^2)/(25x^2-9)dx/sqrt(1-x^2)`

Apply partial fraction decomposition:

`I=-2/3int_0^(1/2)(1/(3+5x)+1/(3-5x))dx-1/5int_0^(1/2)(1-16/(25x^2-9))dx/sqrt(1-x^2)`

Apply the substitution `x=sintheta`:

`I=-2/15[ln|3+5x|-ln|3-5x|]_ 0^(1/2)-1/5int_0^(pi/6)(1-16/(25sin^2theta-9))d theta`

Integration is distributive:

`I=-2/15ln(11)-1/5int_0^(pi/6)d theta+16/5int_0^(pi/6)1/(25sin^2theta-9)d theta`

Rearrange:

`I=-2/15ln(11)-pi/30+16/5int_0^(pi/6)sec^2theta/(25tan^2theta-9sec^2theta)d theta`

Since `sec^2theta=tan^2theta+1`:

`I=-2/15ln(11)-pi/30+16/5int_0^(pi/6)sec^2theta/(16tan^2theta-9)d theta`

Apply the difference of squares:

`I=-2/15ln(11)-pi/30-16/5int_0^(pi/6)sec^2theta/((3+4tantheta)(3-4tantheta))d theta`

Apply partial fraction decomposition:

`I=-2/15ln(11)-pi/30-8/15int_0^(pi/6)(1/(3+4tantheta)+1/(3-4tantheta))sec^2thetad theta`

Integrate term by term:

`I=-2/15ln(11)-pi/30-2/15[ln|3+4tantheta|-ln|3-4tantheta|]_0^(pi/6)`

Insert the limits of integration:

`I=-2/15ln(11)-pi/30-2/15ln((3sqrt3+4)/(3sqrt3-4))`

Rationalize:

`I=-2/15ln(11)-pi/30-2/15ln((3sqrt3+4)^2/11)`

Simplify:

`I=-pi/30-4/15ln(3sqrt3+4)`

Answer 2

`-pi/30+4/15ln|(5-3sqrt3)/(7-3sqrt3)|`.

Explanation:

Let, `I=int_0^(1/2)dx/(4-5sqrt(1-x^2)`.

Subst.ing `x=sinu, dx=cosudu." Also, "sqrt(1-x^2)=cosu`.

When `x=0 rArr sinu=0 rArr u=0`.

When `x=1/2 rArr u=pi/6`.

`:. I=int_0^(pi/6)(cosudu)/(4-5cosu)=-1/5int_0^(pi/6)(-5cosu)/(4-5cosu)du`,

`=-1/5int_0^(pi/6){(4-5cosu)-4}/(4-5cosu)du`,

`=-1/5int_0^(pi/6)(4-5cosu)/(4-5cosu)du+4/5int_0^(pi/6)1/(4-5cosu)du`,

`=-1/5[u]_0^(pi/6)+4/5J," where, "J=int_0^(pi/6)1/(4-5cosu)du`,

`=-1/5[pi/6-0]+4/5J=-pi/30+4/5J`,

For `J," we subst. "tan(u/2)=v. :. sec^2(u/2)*1/2*du=dv`.

`:. du=2*(dv)/sec^2(u/2)=(2dv)/(1+tan^2(u/2))=(2dv)/(1+v^2)...(1)`.

Further, `4-5cosu=4-5{(1-tan^2(u/2))/(1+tan^2(u/2))}`,

`=4-5{(1-v^2)/(1+v^2)}={4(1+v^2)-5(1-v^2)}/(1+v^2)`,

`:. 4-5cosu=(9v^2-1)/(1+v^2).......(2)`.

`"Also, "u=0 rArr v=0, &, u=pi/6, v=tan(pi/12)`.

Utilising `(1) and (2)`, we get,

`J=2int_0^(tan(pi/12))1/{(3v)^2-(1)^2}dv`,

`=2*1/2*1/3[ln|(3v-1)/(3v+1)|]_0^(tan(pi/12))`,

`=1/3ln|(3tan(pi/12)-1)/(3tan(pi/12)+1)|`.

`"Now, "tan(pi/12)=(sqrt3-1)/(sqrt3+1)=(sqrt3-1)^2/(3-1)=2-sqrt3`.

`:. J=1/3ln|{3(2-sqrt3)-1}/{3(2-sqrt3)+1}|=1/3ln|(5-3sqrt3)/(7-3sqrt3)|`.

`:. I=-pi/30+4/15ln|(5-3sqrt3)/(7-3sqrt3)|`.