Question

How do you integrate `int sqrt(5-4x-x^2) dx` using trigonometric substitution?

Answer 1

`((x+2)sqrt(5-4x-x^2)+9arcsin((x+2)/3))/2+C`

Explanation:

We have:

`I=intsqrt(5-4x-x^2)dx=intsqrt(-(x+2)^2+9)dx`

From here, let `3sintheta=x+2`. Thus, `3costhetad theta=dx`. Plugging these in:

`I=3intcosthetasqrt(-9sin^2theta+9)d theta`

Factoring a `sqrt9=3` from the square root:

`I=9intcosthetasqrt(-sin^2theta+1)d theta`

Recall that since `sin^2theta+cos^2theta=1`, we know that `sqrt(-sin^2theta+1)=costheta`.

`I=9intcos^2thetad theta`

Recall that `cos2theta=2cos^2theta-1`. Thus, `cos^2theta=1/2(cos2theta+1)`.

`I=9/2int(cos2theta+1)d theta=9/2intcos2thetad theta+9/2intd theta`

The first integral can be approached with substitution, or just the reverse chain rule. The second is simple:

`I=9/4int2cos2thetad theta+9/2theta=9/4sin2theta+9/2theta`

Using `sin2theta=2sinthetacostheta`:

`I=9/2sinthetacostheta+9/2theta`

Writing all in terms of sine:

`I=9/2sinthetasqrt(1-sin^2theta)+9/2theta`

Recall that `sintheta=(x+2)/3`, so we also see that `theta=arcsin((x+2)/3)`:

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

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

The `sqrt(1/9)=1/3` will come out of the square root and cancel with the `3`:

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

`I=((x+2)sqrt(5-4x-x^2)+9arcsin((x+2)/3))/2+C`