Question

How do you integrate `int sqrt(9-x^2)` using trig substitutions?

Answer 1

`=1/2[9sin^(-1)(x/3) + xsqrt(9-x^2)]+C`

Explanation:

Let `x = 3sin(u) implies dx = 3cos(u)du`

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

`int sqrt(9-x^2)dx = 3int sqrt(9 - 9sin^2(u))cos(u)du`

`= 9int sqrt(1-sin^2(u))cos(u)du = 3int sqrt(cos^2(u))cos(u)du`

`=9int cos^2(u)du`

From double angle formula

`cos2theta = 2cos^2theta - 1 implies cos^2theta = 1/2(1+cos2theta)`

`=9/2int (1 + cos(2u))du`

`= 9/2[u + 1/2sin(2u)] + C`

Apply double angle formula:

`=9/2[u + sin(u)cos(u)] + C`

Rewrite `cos(u) = sqrt(1-sin^2(u))`

`=9/2[u+sin(u)sqrt(1-sin^2(u))]+C`

`=9/2[sin^(-1)(x/3) + x/3sqrt(1-(x/3)^2)]+C`

`=9/2[sin^(-1)(x/3) + x/9sqrt(9-x^2)]+C`

`=1/2[9sin^(-1)(x/3) + xsqrt(9-x^2)]+C`

Answer 2

`9/2arcsin(x/3)+x/2sqrt(9-x^2)+C`.

Explanation:

Let us subst. `x=3sint`, so, `dx=3costdt`

Also, `sqrt(9-x^2)=sqrt(9-9sin^2t)=sqrt(9cos^2t)=3cost`.

Hence, `I=intsqrt(9-x^2)dx`

`=int3cost*3costdt=9intcos^2tdt=9/2int(1+cos2t)dt`

`=9/2{t+sin(2t)/2}=9/2{t+sintcost}`

Here, `x=3sint rArr t=arcsin(x/3)`.

Therefore, `I=9/2{arcsin(x/3)+x/3*sqrt(9-x^2)/3}`

`=9/2arcsin(x/3)+x/2sqrt(9-x^2)+C`.