How do you find the integral of (x+3) *sqrt(4-x^2)(x+3)4x2?

2 Answers

3/2x\sqrt{4-x^2}-1/3(4-x^2)^{3/2}+6\sin^{-1}(x/2)+C32x4x213(4x2)32+6sin1(x2)+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

Jul 27, 2018

-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.