Question

How do you find the integral of `(x+3) *sqrt(4-x^2)`?

Answer 1

`3/2x\sqrt{4-x^2}-1/3(4-x^2)^{3/2}+6\sin^{-1}(x/2)+C`

Explanation:

Let `x=2\sin\theta\implies dx=2\cos\theta\ d\theta`

`\therefore \int (x+3)\sqrt{4-x^2}\ dx`

`=\int (2\sin\theta+3)\sqrt{4-4sin^2\theta}\ (2\cos\theta\ d\theta)`

`=\int (2\sin\theta+3)(2\cos\theta)\ (2\cos\theta\ d\theta)`

`=4\int (2\sin\theta+3)\cos^2\theta\ d\theta`

`=4\int (2\cos^2\theta \sin\theta+3\cos^2\theta)\ d\theta`

`=8\int \cos^2\theta \sin\theta\ \d\ theta+12\int \cos^2\theta\ d\theta`

`=-8\int \cos^2\theta(- \sin\theta\ \d\ theta)+12\int \frac{ 1+\cos2\theta}{2}\ d\theta`

`=-8\int \cos^2\theta \ d(\cos\theta)+6\int (1+\cos2\theta)\ d\theta`

`=-8(\frac{\cos^3\theta}{3})+6(\theta+{\sin2\theta}/2)+C`

`=-8/3\cos^3\theta+6\theta+3\sin2\theta+C`

`=6\sin\theta\cos\theta-8/3\cos^3\theta+6\theta+C`

`=6(x/2)\sqrt{1-(x/2)^2}-8/3(1-(x/2)^2)^{3/2}+6\sin^{-1}(x/2)+C`

`=3/2x\sqrt{4-x^2}-1/3(4-x^2)^{3/2}+6\sin^{-1}(x/2)+C`

Answer 2

`-1/3(4-x^2)^(3/2)+3/2{xsqrt(4-x^2)+4arcsin(x/2)}+C.`

Explanation:

Let us have a Second Solution without using Trigo. Substn.

We will use `intsqrt(a^2-x^2)dx=1/2{xsqrt(a^2-x^2)+a^2arcsin(x/a)}`.

Let, `I=int(x+3)sqrt(4-x^2)dx`.

`:. I=intxsqrt(4-x^2)dx+3intsqrt{2^2-x^2}dx`,

`=I_1+3/2{xsqrt(4-x^2)+4arcsin(x/2)}`, where,

`I_1=intxsqrt(4-x^2)dx`,

`=-1/2intsqrt(4-x^2)(-2xdx)`,

`=-1/2intu^(1/2)du; u=4-x^2, du=-2xdx`,

`=-1/2*u^(1/2+1)/(1/2+1)`,

`=-1/3u^(3/2)`.

`:. I_1=-1/3(4-x^2)^(3/2)`.

Altogether,

`I=-1/3(4-x^2)^(3/2)+3/2{xsqrt(4-x^2)+4arcsin(x/2)}+C.`