# Integral of square root - Conflicting solutions

Can a kind person explain to me why I appear to have two conflicting solutions to:

$$\int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2}$$

Solution 1 : Standard trigonometric substitution: $$x=\sin\theta$$

Integral becomes

$$\int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2}=\int^{\sin^{-1}\frac{1}{\sqrt{2}}}_{\sin^{-1}0}\frac{d\theta}{\cos\theta}\sqrt{1-\sin^2\theta}=\int^{\pi/4}_0d\theta=\frac{\pi}{4}$$

Solution 2 : Integration by parts gives:

$$\int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2}=\left.x\sqrt{1-x^2}\right|^{\frac{1}{\sqrt{2}}}_0 + \int^{\frac{1}{\sqrt{2}}}_0dx\frac{x^2}{\sqrt{1-x^2}}=\frac{1}{2}+\int^{\frac{1}{\sqrt{2}}}_0dx \frac{x^2}{\sqrt{1-x^2}}$$

Adding to the right hand side:

$$0=\int^{\frac{1}{\sqrt{2}}}_0dx\frac{1}{\sqrt{1-x^2}}-\int^{\frac{1}{\sqrt{2}}}_0dx\frac{1}{\sqrt{1-x^2}}$$

we get:

$$\int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2}=\frac{1}{2} - \int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2}+\int^{\frac{1}{\sqrt{2}}}_0dx\frac{1}{\sqrt{1-x^2}}$$

or rearranging:

$$\int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2}=\frac{1}{4}+\frac{1}{2}\int^{\frac{1}{\sqrt{2}}}_0dx \frac{1}{\sqrt{1-x^2}}$$

Finally since:

$$\sin^{-1} x=\int^x_0\frac{dz}{\sqrt{1-z^2}}\quad |x|\leq1$$

we find:

$$\int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2}=\frac{1}{4}+\frac{1}{2}\sin^{-1}\frac{1}{\sqrt{2}}=\frac{1}{4}+\frac{\pi}{8}$$

Any illuminating comments are appreciated.

Regards,
Beauty

## Answers and Replies

Ok forget it, my first integral is completely wrong.