Question

How do you integrate `int x^2/sqrt(x^2-81)dx` using trigonometric substitution?

Answer 1

`I=x/2sqrt(x^2-81)+81/2ln|x+sqrt(x^2-81)|+C`

Explanation:

Here,

`I=intx^2/sqrt(x^2-81)dx`

Subst. `x=9secu=>dx=9secutanudu, and secu=x/9`

So,

`I=int (81sec^2u)/sqrt(81sec^2u-81)xx9secutanudu`

`=int(81sec^2u)/(cancel9canceltanu)xxcancel9secucanceltanudu`

`=81intsecu*sec^2udu`

`=81intsqrt(1+tan^2u)*sec^2udu`

Let, `tanu=t=>sec^2udu=dt`

`:.I=81intsqrt(1+t^2)dt`

`=81[t/2sqrt(1+t^2)+1^2/2ln|t+sqrt(1+t^2)|]+c`

Subst. back , `t=tanu`

`I`=`81[tanu/2sqrt(1+tan^2u)+1/2ln|tanu+sqrt(1+tan^2u)|]+C'`

=`81[sqrt(sec^2u-1)/2xxsecu+1/2ln|sqrt(sec^2u-1)+secu|]+C'`

Again subst. `secu=x/9`

`I=81[sqrt((x^2)/81-1))/2xxx/9+1/2ln|sqrt((x^2/81)-1)+x/9|]+C'`

`=81[sqrt(x^2-81)/(2xx9)*x/9+1/2ln|sqrt(x^2-81)/9+x/9|]+C'`

`=x/2sqrt(x^2-81)+81/2ln|(x+sqrt(x^2-81))/9|+C'`
`=x/2sqrt(x^2-81)+81/2[ln|x+sqrt(x^2-81)|-ln9]+C'`

`=x/2sqrt(x^2-81)+81/2ln|x+sqrt(x^2-81)|-81/2ln9+C'`

`=x/2sqrt(x^2-81)+81/2ln|x+sqrt(x^2-81)|+C`

where `C=C'-81/2ln9`

Answer 2

`x/2sqrt(x^2-81)+81/2ln|(x+sqrt(x^2-81))|+C.`

Explanation:

Kindly refer to the Second Method which does not use the

Trigo. Subst. to solve the Problem.

Prerequisite :

`intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-a^2/2ln|(x+sqrt(x^2-a^2))|+c_1`,

&, `intdx/sqrt(x^2-a^2)=ln|(x+sqrt(x^2-a^2))|+c_2`,

Suppose that, `I=intx^2/sqrt(x^2-81)dx`.

`:. I=int{(x^2-81)+81}/sqrt(x^2-81)dx`,

`=int{(x^2-81)/sqrt(x^2-81)+81/sqrt(x^2-81)}dx`,

`=intsqrt(x^2-81)dx+81intdx/sqrt(x^2-81)`,

`=x/2sqrt(x^2-81)-81/2ln|(x+sqrt(x^2-81)|`

`+81ln|(x+sqrt(x^2-81)|`.

`:. I=x/2sqrt(x^2-81)+81/2ln|(x+sqrt(x^2-81))|+C`, as

Respected Maganbhai P. has already derived!.

Enjoy Maths.!