How to integrate ∫ √(4 - x²) dx?

Question

I've never done an integral like this, but this is what I've got so far:
Since the integral is in the form ∫ √(a² - x²) dx, i used the substitution x = asinθ, in this case
x = 2sinθ, so dx = 2 cosθ dθ.
When you plug it in you get:
∫ √(4 - (2sinθ)²) (2 cosθ) dθ =
2 ∫ √(4 - 4sin²θ) cosθ dθ =
2 ∫ √(4 (1 - sin²θ)) cosθ dθ =
2 ∫ 2√(cos²θ) cosθ dθ =
4 ∫ cos²θ dθ =
4 (1/2*θ + 1/4*sin(2θ) ) =

2θ + sin(2θ) + C

The problem is converting this answer back to x, which my book does not really discuss.
Since x = 2 sin(θ), θ = arcsin(x/2)
So 2θ + sin(2θ) + C becomes
2*arcsin(x/2) + sin(2θ) + C.
But how do you convert the sin(2θ)?
I tried the double angle formula and it didn't work well. The answer in my textbook is:
2*arcsin(x/2) + 1/2*x√(4 - x²) + C
I would greatly appreciate any help.

thanks dr_no4458. My book left out the details on how to do this using a right triangle. Now it makes perfect sense.

Answer 1

Everything looks good but to convert sin 2theta, draw a triangle. you know that x = 2 *sin(theta). So the triangle would have the hypotenuse of 2 and the opposite side is x. So using the pythagorean theorem you find that adjacent side is sqrt(4- x^2).
Your final answer will then be:
2(arcsin(x/2)) +2sin(theta)cos(theta) + C =
2(arcsin(x/2)) + 2* (x/2)(sqrt(4- x^2)) + C
The key here is to draw the triangle relating x and theta.

Answer 2

Your problem is how
sin 2θ became 1/2*x√(4-x²) ;isn't it?

sin 2θ = 2sin θcos θ.
But 2Sin θ = x;
Now draw a triangle taking an angle θ, mark
the opposite side as x and the hypotenuse as 2
{As Sin θ = x/2}
Adjacent side = √(4-x²).

cos θ = (√(4-x²))/2;

Now,
substitute the values of sin θ and cos θ in the equation
sin 2θ = 2sin θcos θ.
You will get
sin 2θ = 1/2*x√(4 - x²)

Answer 3

sin(2T) = 2sin(T)cos(T)

now T = arcsin(x/2) so sin(T) = x/2 and cos(T)=SQRT(4-x^2)/2

Draw a right triangle. Put C=2 and either A or B =2 (I will use A)
and the angle T is opposite the side A
So the other side of the triangle is B = SQRT(C^2 - A^2)
Giving B = SQRT(4 - x^2) and cos(T) = SQRT(4 - x^2)/2

then sin(2T) = 2*(x/2)*SQRT(4 - x^2)/2 = SQRT(4 - x^2)/2

Answer 4

You've done everything right so far. The trick is then just to convert sin 2theta into something in terms of x.

Remember your trig identities:
sin2 theta = 2 sin theta cos theta, and
cos theta = sqrt(1-sin^2 theta).

With these two points, you can convert sin 2 theta into 1/2 x sqrt(4-x^2) in a very straightforward way.

Answer 5

Thr is a direct formula for such questions.U pleez go thru yr books and u shud find the formula.