- #1
BeautyT
- 2
- 0
Can a kind person explain to me why I appear to have two conflicting solutions to:
[tex] \int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2} [/tex]
Solution 1 : Standard trigonometric substitution: [tex] x=\sin\theta [/tex]
Integral becomes
[tex] \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} [/tex]
Solution 2 : Integration by parts gives:
[tex] \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}} [/tex]
Adding to the right hand side:
[tex] 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}} [/tex]
we get:
[tex] \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}} [/tex]
or rearranging:
[tex] \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}} [/tex]
Finally since:
[tex] \sin^{-1} x=\int^x_0\frac{dz}{\sqrt{1-z^2}}\quad |x|\leq1 [/tex]
we find:
[tex] \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} [/tex]
Any illuminating comments are appreciated.
Regards,
Beauty
[tex] \int^{\frac{1}{\sqrt{2}}}_0dx\sqrt{1-x^2} [/tex]
Solution 1 : Standard trigonometric substitution: [tex] x=\sin\theta [/tex]
Integral becomes
[tex] \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} [/tex]
Solution 2 : Integration by parts gives:
[tex] \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}} [/tex]
Adding to the right hand side:
[tex] 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}} [/tex]
we get:
[tex] \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}} [/tex]
or rearranging:
[tex] \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}} [/tex]
Finally since:
[tex] \sin^{-1} x=\int^x_0\frac{dz}{\sqrt{1-z^2}}\quad |x|\leq1 [/tex]
we find:
[tex] \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} [/tex]
Any illuminating comments are appreciated.
Regards,
Beauty