What value of limits should I substitute in the arcsin(x/a) term in the formula for ∫(4-x²)½ dx where x goes from -1 to 1

Question

Logically the area would be half a circle=2π

Ok so, we have this formula. The only problem is the arcsin term gives π/6,5π/6,-7π/6,-11π/6 and their nπ multiples for x=1 in the formula. A similar situation holds for the arcsin term with x=-1 Which value of angles do we use to solve this and arrive to the answer=2π

I know the curve is in the first and second quadrant. I tried substituting the angles related to the same but with no success

Edit: I made an error x should go from -2 to 2 and the angles I mentioned above was for arcsin(1/2) and it should be arcsin(2/2) but the question still holds up to which value of angles should I substitute for arcsin(1/2) from π/6,5π/6,-7π/6,-11π/6 and their nπ multiples ? Similar things go for arcsin(-1/2). Do all the angles chosen have to be from the positive axis? I'm seeing answers where arcsin(1/2)=π/6 and arcsin(-1/2)=-π/6. Shouldn't we be using 11π/6 instead of -π/6 so all the angles are from the postive x axis?

Answer 1

First of all the bounds for $x$ are $\pm 2$

Secondly you have $$\sin ^{-1}(x/2)|_{-2}^{2} = \sin^{-1} (1) - \sin ^{-1}(-1)= \pi/2 -(-\pi/2) = \pi $$

Answer 2

We have that

• $\arcsin(-1)=-\frac{\pi}2$

• $\arcsin(1)=\frac{\pi}2$

the added angle is not important since it cancels out for the definite integral.

Answer 3

Use $a = 2$ in the evaluation below,

$$I = \int_{-1}^1 \sqrt{4-x^2}dx = \frac x2 \sqrt{4-x^2}|_{-1}^1 + 2\sin^{-1}\frac x2 |_{-1}^1$$ $$= \frac 12[\sqrt 3-(-\sqrt 3)]+2\left[\frac{\pi}{6}-(-\frac{\pi}{6})\right]=\sqrt 3+\frac 23\pi$$

The two angles are to be chosen corresponding to the variable substitution $x=2\sin t$ in carrying out the integral. The inverse function $\sin^{-1}(t)$ is only defined in the domain [-1,1] with the range $(-\frac{\pi}{2},\frac{\pi}{2})$. So, they are $\sin^{-1}(\pm \frac 12) = \pm \frac{\pi}{6}$.

The two angles are not related to the polar coordinates or quadrant locations; they result from the variable substitution in integrations.

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Note: The integral does not give the area of a half circle. It is smaller since the integration range is (-1,1). The half circle integral is, instead of

$$I = \int_{-2}^2 \sqrt{4-x^2}dx$$

$$=\frac x2 \sqrt{4-x^2}|_{-2}^2 + 2\sin^{-1}\frac x2 |_{-2}^2= 2\left[\frac{\pi}{2}-(-\frac{\pi}{2})\right]=2\pi$$