Question

How do you integrate `int x^2/(sqrt(x^2-4))dx` using trigonometric substitution?

Answer 1

`x/2sqrt(x^2-4)+2ln|x/2+sqrt(x^2-4)/2|+C, or,`

`x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+c, c=C-2ln2`.

Explanation:

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

We subst. `x=2secy," so that, "dx=2secytanydy`.

`:. I=int(4sec^2y)/sqrt(4sec^2y-4)*2secytanydy`,

`=int(4sec^2y)/(2tany)*2secytanydy`,

`=4intsec^3ydy`,

`=4intsecy*sec^2ydy`,

`=4intsqrt(tan^2y+1)sec^2ydy`.

Next, we use the substn. `tany=t :. sec^2ydy=dt`.

`:. I=4intsqrt(t^2+1)dt`,

`=4{t/2sqrt(t^2+1)+1/2ln|t+sqrt(t^2+1)|}`,

`=2{tanysecy+ln|tany+secy|}......[because, t=tany]`.

Returning to `secy=x/2," so that, "tany=sqrt(x^2/4-1)`, we have,

`I=2{x/2*sqrt(x^2-4)/2+ln|x/2+sqrt(x^2-4)/2|}`,

`rArr I=x/2sqrt(x^2-4)+2ln|x/2+sqrt(x^2-4)/2|+C, or,`

`I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+c, c=C-2ln2`.

However, the integral can easily be dealt with without

using the substn. as shown below :

`I=intx^2/sqrt(x^2-4)dx=int{(x^2-4)+4}/sqrt(x^2-4)dx`,

`int{(x^2-4)/sqrt(x^2-4)+4/sqrt(x^2-4)}dx`,

`=intsqrt(x^2-4)dx+4int1/sqrt(x^2-4)dx`,

`={x/2sqrt(x^2-4)-4/2ln|x+sqrt(x^2-4)|}+4ln|x+sqrt(x^2-4)|`.

`I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+C_1`, as before!

Enjoy Maths.!

Answer 2

`I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+c`
If we take `x=2sectheta` leads to `I=intsec^3thetad(theta)=intsqrt(1+tan^2theta)sec^2thetad(theta)`....Again take `tantheta=t=>I=intsqrt(1-t^2)dt`,then use (1)

Explanation:

We have,
`color(red)((1)intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-(a^2)/2ln|x+sqrt(x^2-a^2)|`
`color(red)((2)int1/sqrt(x^2-a^2)dx=ln|x+sqrt(x^2-4)|+c`
Hence,
`I=int(x^2)/sqrt(x^2-4)dx=int((x^2-4)+4)/sqrt(x^2-4)dx`
`I=int(x^2-4)/sqrt(x^2-4)dx+int4/sqrt(x^2-4)dx`
`=intsqrt(x^2-4) dx+4int1/sqrt(x^2-4)dx`
`=intsqrt(x^2-2^2)dx+4int1/sqrt(x^2-2^2)dx`
Using (1) and (2)
`I=x/2sqrt(x^2-2^2)-(2^2)/2ln|x+sqrt(x^2-2^2)|+4ln|x+sqrt(x^2-2^2)|`
`=x/2sqrt(x^2-4)-2ln|x+sqrt(x^2-4)|+4ln|x+sqrt(x^2-4)|+c`
`I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+c`

For Trigonometric substitution proceed according to Hint given
above