Integration of √(1-x^2) with x=sinθ: Check Correctness and Simplification

[askor]

Does my below integration is correct?

$\int \sqrt{1 - x^2} \ dx$

Let $x = \sin \theta$, then
$dx = \cos \theta \ d \theta$,
$\cos \theta = \sqrt{1 - x^2}$,
$\theta = \sin^{-1} (x)$

$\int \sqrt{1 - x^2} \ dx$
$= \int \left( \sqrt{1 - \sin^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)$
$= \int \left( \sqrt{\cos^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)$
$= \int \cos \theta \cos \theta \ d\theta$
$= \int \cos^2 \theta \ d\theta$
$= \int \frac{1 + \cos 2\theta}{2} \ d\theta$
$= \frac{1}{2} \int d\theta + \frac{1}{2} \int \cos 2\theta \ d\theta$
$= \frac{1}{2} \theta + \frac{1}{2} \frac{1}{2} \sin 2\theta + C$
$= \frac{1}{2} \theta + \frac{1}{4} \sin 2\theta + C$
$= \frac{1}{2} \theta + \frac{1}{4} 2 \sin \theta \cos \theta + C$
$= \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta + C$
$= \frac{1}{2} \sin^{-1} x + \frac{1}{2} x \sqrt{1 - x^2} + C$

Does this correct?

 

[fresh_42]

Does my below integration is correct?

$\int \sqrt{1 - x^2} \ dx$

Let $x = \sin \theta$, then
$dx = \cos \theta \ d \theta$,
$\cos \theta = \sqrt{1 - x^2}$,
$\theta = \sin^{-1} (x)$

$\int \sqrt{1 - x^2} \ dx$
$= \int \left( \sqrt{1 - \sin^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)$
$= \int \left( \sqrt{\cos^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)$
$= \int \cos \theta \cos \theta \ d\theta$
$= \int \cos^2 \theta \ d\theta$
$= \int \frac{1 + \cos 2\theta}{2} \ d\theta$
$= \frac{1}{2} \int d\theta + \frac{1}{2} \int \cos 2\theta \ d\theta$
$= \frac{1}{2} \theta + \frac{1}{2} \frac{1}{2} \sin 2\theta + C$
$= \frac{1}{2} \theta + \frac{1}{4} \sin 2\theta + C$
$= \frac{1}{2} \theta + \frac{1}{4} 2 \sin \theta \cos \theta + C$
$= \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta + C$
$= \frac{1}{2} \sin^{-1} x + \frac{1}{2} x \sqrt{1 - x^2} + C$

Does this correct?

Yes, in case $|x|\leq 1.$

 

[askor]

Yes, in case $|x|\leq 1.$

Why $|x|\leq 1.$?

 

[fresh_42]

Why $|x|\leq 1.$?

Otherwise, it won't be the sine of an angle.

The formula for $|x|\geq 1$ involves $\cosh^{-1}$ at least if the integral table on Wikipedia where I looked it up is correct.

 

[pasmith]

If $x > 1$ the integral is $i\int \sqrt{x^2 - 1}\,dx$ and $x = \cosh u$ with $dx = \sinh u\,du$ and $\cosh^2 u - 1 =\sinh^2 u$ is simplest. If $x < -1$ then use $x = -\cosh u$ instead.

 

[topsquark]

Does my below integration is correct?

$\int \sqrt{1 - x^2} \ dx$

Let $x = \sin \theta$, then
$dx = \cos \theta \ d \theta$,
$\cos \theta = \sqrt{1 - x^2}$,
$\theta = \sin^{-1} (x)$

$\int \sqrt{1 - x^2} \ dx$
$= \int \left( \sqrt{1 - \sin^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)$
$= \int \left( \sqrt{\cos^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)$
$= \int \cos \theta \cos \theta \ d\theta$
$= \int \cos^2 \theta \ d\theta$
$= \int \frac{1 + \cos 2\theta}{2} \ d\theta$
$= \frac{1}{2} \int d\theta + \frac{1}{2} \int \cos 2\theta \ d\theta$
$= \frac{1}{2} \theta + \frac{1}{2} \frac{1}{2} \sin 2\theta + C$
$= \frac{1}{2} \theta + \frac{1}{4} \sin 2\theta + C$
$= \frac{1}{2} \theta + \frac{1}{4} 2 \sin \theta \cos \theta + C$
$= \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta + C$
$= \frac{1}{2} \sin^{-1} x + \frac{1}{2} x \sqrt{1 - x^2} + C$

Does this correct?

One way to find out is to take the derivative. What does that give you?

-Dan

 

[Mark44]

Does this correct?

One way to find out is to take the derivative. What does that give you?

@askor, this is something you should always do in an integration problem. If you differentiate your answer and get back to the original integrand, you know your work is correct (assuming you didn't make any mistakes in differentiating the answer).

 

[askor]

Can I modify the result such as $\theta = \cos^{-1} (\sqrt{1 - x^2})$,
so that the final result will be $\frac{1}{2} \cos^{-1} (\sqrt{1 - x^2}) + \frac{1}{2} x \sqrt{1 - x^2} + C$?

 

[askor]

Hello, anybody home? Can someone please answer my question?

 

[fresh_42]

Hello, anybody home? Can someone please answer my question?

Can I modify the result such as $\theta = \cos^{-1} (\sqrt{1 - x^2})$,
so that the final result will be $\frac{1}{2} \cos^{-1} (\sqrt{1 - x^2}) + \frac{1}{2} x \sqrt{1 - x^2} + C$?

@askor, this is something you should always do in an integration problem. If you differentiate your answer and get back to the original integrand, you know your work is correct (assuming you didn't make any mistakes in differentiating the answer).

https://www.wolframalpha.com/input?i=d/dx+((1/2)*(cos^{-1}+(root(1+-+x^2)+))+++(1/2)+*x+*root(1-x^2))=

 

[askor]

What does it mean? I don't understand.

 

[fresh_42]

What does what mean?
What don't you understand?
Have you differentiated your suggestion as recommended?
What did you mean in post #8?
Where are your calculations?
Do you know WolframAlpha?
Did you click the link?
What from the link didn't you understand?

Listen, you cannot come here and expect people to do your work or to guess what you might have meant.

 

[Fred Wright]

Observe that for $x \gt 1$ the integrand is complex.
$$
I=\int \sqrt{1-x^2}dx=i\int \sqrt{x^2-1}dx
$$
If you want to go further with this; let $x=\cosh(u)$, $dx=\sinh(u)du$.
$$
\sqrt{\cosh^2(u) -1}=\sinh(u)
$$
$$
I=i\int \sinh^2(u)du
$$
$$
\sinh^2(u)=\frac{1}{2}(\cosh(2u)-1)
$$
$$
I=\frac{i}{2}\int (\cosh(2u)-1)du=\frac{i}{4}(\sinh(2u)-2u) + C
$$
$$
=\frac{i}{4}(\sinh(2\cosh^{-1}(x)) -2\cosh^{-1}(x)) + C
$$
$$
=\frac{i}{4}(\sinh(2\ln(x+\sqrt{x^2 -1})) -2\ln(x+\sqrt{x^2 -1})) + C
$$
I advise you not to be so petulant when people are trying to help you.