Question

How to solve solve integral 1/(x+√(9-x^2)) with lower limit 0 and upper limit 3?

Answer 1

`I=pi/4`

Explanation:

We know that,

"If `f` is continous in `[0,a] ,then`

`color(red)(int_0^af(x)dx=int_0^af(a-x)dx...to(F)`"

Here,

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

Let,

`x=3sinu=>dx=3cosudu`

`andsinu=x/3=>u=sin^-1(x/3)`

`x=0=>u=0andx=3=>u=pi/2`

So,

`I=int_0^(pi/2) 1/(3sinu+sqrt(9-9sin^2u))xx3cosudu`

`=int_0^(pi/2) (3cosu)/(3sinu+3cosu)du`

`color(blue)(I=int_0^(pi/2)cosu/(sinu+cosu)du...to(A)`

`I=int_0^(pi/2)(cos(pi/2-u))/((sin(pi/2-u)+cos(pi/2-u)))du...tocolor(red)(Apply(F)`

`color(blue)(I=int_0^(pi/2)(sinu)/(cosu+sinu)du...to(B)`

Adding `(A)and(B)`

`I+I=int_0^(pi/2)[cosu/(sinu+cosu)+sinu/(cosu+sinu)]du`

`2I=int_0^(pi/2)[(cosu+sinu)/(sinu+cosu)]du`

`2I=int_0^(pi/2)1du`

`2I=[u]_0^(pi/2)`

`2I=pi/2-0=pi/2`

`I=pi/4`

Answer 2

`pi/4`.

Explanation:

`"Prerequisite :" int_0^a f(t)dt=int_0^af(a-t)dt............(ast)`.

Let, `I=int_0^3 1/(x+sqrt(9-x^2))dx`.

We subst. `x=3sint. :. dx=3costdt`.

Also, `x=0 rArr t=0; &, x=3 rArr t=pi/2`.

`:. I=int_0^(pi/2) 1/(3sint+3cost)*3costdt`.

`:. I=int_0^(pi/2) cost/(sint+cost)dt.................(1)`.

Applying `(ast)" on "I`, we get,

`I=int_0^(pi/2) cos(pi/2-t)/{sin(pi/2-t)+cos(pi/2-t)}dt, i.e.,`

`I=int_0^(pi/2) sint/(cost+sint)dt.........................(2)`.

`"Adding "(1) & (2), 2I=int_0^(pi/2){(cost+sint)/(cost+sint)}dt`,

`=int_0^(pi/2)1dt`,

`=[t]_0^(pi/2)`.

`rArr 2I=pi/2`.

`rArr I=pi/4`, as desired!

Enjoy Maths!