Trouble solving $\int\sqrt{1-x^2} \, dx$ I am trying to learn how to solve integrals and I've got the hang out of a lot of examples, but I haven't got the slightest idea how to solve this example, this is how far I've got:
$$
\int\sqrt{1 - x^2} \, dx = x\sqrt{1-x^2} - 2\int\frac{x^2}{\sqrt{1-x^2}} \, dx
$$
Can you please help me solve it, and also some tips concerning the integration are welcome.
Thank you
 A: Using Trigonometric substitution, $\arcsin x=\phi$
$\implies (i)x=\sin\phi$ and
$(ii)-\frac\pi2\le \phi\le\frac\pi2$ based on principal value of inverse sine ratio,
$\implies\cos\phi\ge0$
$$\int\sqrt{1-x^2}dx=\int|\cos\phi|\cos\phi d\phi = \int\cos^2\phi d\phi$$
$$= \frac12\int(1+\cos2\phi) d\phi$$
$$=\frac12\left(\phi+\frac{\sin2\phi}2\right)+K=\frac{\phi+\sin\phi\cos\phi}2+K$$
A: $\newcommand{\cos}{\sin}$
Let's try $$\int_0^{x}\sqrt{1-a^2}\text da$$ You can generalize from that.  Now, $\sqrt{1-x^2}$ describes a semicircle.  Consider the following image:
So we have $\int_0^x\sqrt{1-a^2}\text da=A_1+A_2$.  $A_2$ is a right triangle with base $x$ and height $\sqrt{1-x^2},$ so $A_2=\frac12x\sqrt{1-x^2}$.  Meanwhile, $A_2$ is a sector of the circle.  Note that the angle $$\theta=\angle (0,1)O(x,\sqrt{1-x^2})=\frac{\sin^{-1}x}1=\sin^{-1}x$$
Since the angle from the center is proportional to the area, $A_2=\frac{\theta}{2\pi}\cdot \pi r^2=\frac{\sin^{-1}x}{2}$
Thus $$ \int_0^x\sqrt{1-a^2}\text da=\frac{\sin^{-1}x+x\sqrt{1-x^2}}{2} $$
A: subsititute $ x =sin\theta $, this gives, 
$ dx = cos\theta d\theta$
hence, $ \int \sqrt{(1-x^2)} dx  = \int (\sqrt{1 - sin^2\theta} ). cos\theta d\theta = \int cos^2 \theta d\theta $
Remember that, $cos(\alpha + \beta) =  cos \alpha .cos \beta - sin \alpha . sin \beta $, 
substituting, $\alpha = \beta = \theta$, 


*

*$ (cos^2\theta - sin^2\theta) = cos(2\theta)$, as well

*$ (cos^2\theta + sin^2\theta) = 1  $
Adding equation 1. and 2. yields, 
$2 cos^2 \theta = 1 + cos2\theta ~ \implies ~ cos^2\theta = \frac{1}{2}(1+cos2\theta)$
hence going  back to integral,
$\int cos^2\theta d\theta = \frac{1}{2}\int (1+cos2\theta) d\theta  = \frac{1}{2}(\theta + \frac{sin2\theta}{2}) + C$ 
By reverse substitution, $\theta = cos^{-1}x = arccos(x) $  and $sin2\theta = 2.sin\theta . cos\theta$
The solution of indefinite integral is ,
$ \int \sqrt{(1-x^2)} dx =  \frac{1}{2} ( \theta + \frac{2.sin\theta.cost\theta}{2} )+ C \implies \frac{1}{2}(\theta + sin\theta. cos\theta) + C $
$ \int \sqrt{(1-x^2)} dx \implies  \frac{1}{2}(arccos(x) + x\sqrt{1-x^2}) + C $
A: I, personally, suggest @lab's way because it is general and suitable for students, but you may use this strange substitution as well: $$1-x^2=x^2t^2$$
A: Your idea of integrating by parts is good (but you did it wrong):
\begin{align}
\int\sqrt{1 - x^2} \, dx 
&= x\sqrt{1-x^2} - \int x\frac{-2x}{2\sqrt{1-x^2}} \, dx \\
&= x\sqrt{1-x^2} + \int\frac{x^2}{\sqrt{1-x^2}}\,dx\\
&= x\sqrt{1-x^2} + \int\frac{x^2-1+1}{\sqrt{1-x^2}}\,dx\\
&= x\sqrt{1-x^2} - \int\sqrt{1-x^2}\,dx+\int\frac{1}{\sqrt{1-x^2}}\,dx\\
&= x\sqrt{1-x^2} + \arcsin x - \int\sqrt{1-x^2}\,dx
\end{align}
Therefore
$$
2\int\sqrt{1 - x^2} \, dx = x\sqrt{1-x^2} + \arcsin x
$$
and
$$
\int\sqrt{1 - x^2} \, dx = \frac{1}{2}x\sqrt{1-x^2} + \frac{1}{2}\arcsin x + c
$$
