Integral of $\int \sqrt{1-4x^2}$ I know I am messing up something with the substitutions but I am not sure what.
$$\int \sqrt{1-4x^2}$$
$$u = 4x, du = 4 \,dx$$
$$\frac{1}{4}\int \sqrt{1-u^2}$$
$u = \sin \theta$ 
$$\frac{1}{4}\int \sqrt{1-\sin^2 \theta} = \frac{1}{4}\int \sqrt{ \cos^2 \theta} = \frac{1}{4}\int \cos \theta = \frac{\sin \theta}{4}$$
Replace $\theta$ with $u$
$$\frac{\sin (\arcsin u)}{4} = \frac{u}{4} = \frac{4x}{4} = x$$
This is wrong and I have no idea why.
 A: $$\int \sqrt{1-4x^2}\ dx$$
Substitute
$$\begin{align*}x &= \frac{\sin (\theta)}{2}\\ dx &= \frac{\cos(\theta)}{2}d \theta\end{align*}$$
So the intregral become
$$\begin{align*}
\int \sqrt{1-4x^2}\ dx &=\frac{1}{2}\int \cos^2(\theta)d\theta\\
&=\frac{1}{2}\int \frac{\cos(2\theta) + 1}{2}d\theta\\
&=\frac{\sin(2\theta)}{8}+\frac{\theta}{4}
\end{align*}$$
Again 
$$\theta = \arcsin(2x)$$
And
$$\begin{align*}
\sin 2\theta &= 2\sin \theta \cos \theta\\
             &=4x\sqrt{1-4x^2}
\end{align*}$$
So the answer is
$$\frac{x\sqrt{1-4x^2}}{2}+\frac{\arcsin(2x)}{4}+C$$
A: You should substitute $u=2x$, to get the next integral:
$$\int du/2 \sqrt{1-u^2}$$
Now change variables to $u=\cos(t)$ to get:
$$-\int (\sin(t)/2) \sin(t) dt$$ which you solve by parts or by using the formula for $\cos(2\theta)$.
Hope I helped.
A: Some things to watch out for:


*

*As already mentioned, if you substitute $u = 4x$, then you get $u^2 = (4x)^2 = 16x^2$ instead of $u^2 = 4x^2$. If you use $u = 2x$, then $u^2 = (2x)^2 = 4x^2$.

*An annoying but important part of practicing good mathematics is being precise. 


*

*In your integrals, you forgot the $dx$ and $du$ several times, which also lead you to forget to express $du$ in terms of $d\theta$. Perhaps you have heard of Riemann integrals; these may help you understand why the "$dx$" is there.

*You forgot the fact that if $F'(x) = f(x)$, then $\int f(x) dx = F(x) + C$. After all, when taking derivatives, all constants disappear, so the derivatives of $F(x)$ and $F(x) + 5$ are exactly the same. Thus, when calculating indefinite integrals, you do not know the constant term of the resulting function. So we generally write $\int f(x) dx = F(x) + C$ for some unknown constant $C$.



Here's a step-by-step approach of how I would do it. First, substitute $u = 2x$, so that $u^2 = (2x)^2 = 4x^2$, and apply $du = 2dx$:
$$\int \sqrt{1 - \color{red}{4x^2}} \color{blue}{dx} = \int \sqrt{1 - \color{red}{u^2}} \color{blue}{\frac{du}{2}} = \frac{1}{2} \int \sqrt{1 - u^2} du.$$
Then substitute $u = \sin \theta$ and $du = \cos \theta d\theta$ to get:
$$\frac{1}{2} \int \sqrt{1 - \color{red}{u^2}} \color{blue}{du} = \frac{1}{2} \int \sqrt{1 - \color{red}{\sin^2 \theta}} \color{blue}{\cos \theta d\theta} = \frac{1}{2} \int \sqrt{\cos^2 \theta} \cos \theta d\theta = \frac{1}{2} \int \cos^2 \theta d\theta.$$
Next, apply the double angle formula for the cosine, $\cos^2 \theta = \frac{1}{2} (1 + \cos 2\theta)$, to get
$$\frac{1}{2} \int \color{red}{\cos^2 \theta} d\theta = \frac{1}{2} \int \color{red}{\frac{1 + \cos 2\theta}{2}} d\theta = \frac{1}{4} \int (1 + \cos 2\theta) d\theta.$$
Now split in two integrals and calculate those separately:
$$\frac{1}{4} \int (\color{red}{1} + \color{blue}{\cos 2\theta}) d\theta = \frac{1}{4} \left(\int \color{red}{1} d\theta + \int \color{blue}{\cos 2\theta} d\theta\right) = \frac{1}{4}\left(\color{red}{\theta + C_1} + \color{blue}{\frac{\sin 2\theta}{2} + C_2}\right) \\ = \frac{\theta}{4} + \frac{\sin 2\theta}{8} + C.$$
Applying a double angle formula for the sine, $\sin 2\theta = 2 \sin \theta \cos \theta$, we get
$$\frac{\theta}{4} + \frac{\color{red}{\sin 2\theta}}{8} + C = \frac{\theta}{4} + \frac{\color{red}{2 \sin \theta \cos \theta}}{8} + C = \frac{\theta}{4} + \frac{\sin \theta \cos \theta}{4} + C.$$
Substituting back $\theta = \arcsin u$, and using $\cos (\arcsin u) = \sqrt{1 - u^2}$, we get
$$\frac{\color{red}{\theta}}{4} + \frac{\sin \color{blue}{\theta} \cos \color{green}{\theta}}{4} + C = \frac{\color{red}{\arcsin u}}{4} + \frac{\sin(\color{blue}{\arcsin u}) \cos(\color{green}{\arcsin u})}{4} + C = \frac{\arcsin u}{4} + \frac{u \sqrt{1 - u^2}}{4} + C.$$
Finally substituting $u = 2x$, you get
$$\frac{\arcsin \color{red}{u}}{4} + \frac{\color{blue}{u} \sqrt{1 - \color{green}{u}^2}}{4} + C = \frac{\arcsin \color{red}{2x}}{4} + \frac{\color{blue}{2x} \sqrt{1 - \color{green}{(2x)}^2}}{4} + C = \boxed{\displaystyle\frac{\arcsin 2x}{4} + \frac{x \sqrt{1 - 4x^2}}{2} + C}.$$
A: Let $2x = u$. So, $dx = \frac{1}{2}du$
$$\frac{1}{2}\int \sqrt{1-u^2} du$$
Replace $u = \sin \theta$, $du = cos \theta\ d\theta$
So, equation will be
$$\frac{1}{2}\int \sqrt{1-sin^2 \theta}\ cos\theta\ d\theta$$
$$\frac{1}{2}\int \sqrt{\cos^2\theta}\ cos\theta\ d\theta$$
$$\frac{1}{2}\int  cos^2\theta\ d\theta$$
Add and subtract $\frac{1}{4}\int d\theta$
$$\frac{1}{4}\int  2cos^2\theta\ d\theta - \frac{1}{4}\int d\theta + \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int  (2cos^2\theta-1)\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int  (2cos^2\theta-1)\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{4}\int  \cos2\theta\ d\theta+ \frac{1}{4}\int d\theta$$
$$\frac{1}{8}  (\sin2\theta) + \frac{1}{4} \theta + C$$
$$\frac{1}{8} (2\sin\theta.\cos\theta) + \frac{1}{4} \theta + C$$
As u = $\sin\theta$
$$\frac{1}{4} (u\sqrt{1-u^2}) + \frac{1}{4} \sin^{-1}u + C$$
As $2x = u$
$$\frac{1}{4} (2x\sqrt{1-4x^2}) + \frac{1}{4} \sin^{-1}2x + C$$
$$\frac{1}{2} (x\sqrt{1-4x^2}) + \frac{1}{4} \sin^{-1}2x + C$$
