1. ## Trig Substitution

Where is my error in this problem?

$\int\sqrt{1-4x^2}\,dx$

$x=\frac{1}{2}\sin{t}$
$dx=\frac{1}{2}\cos{5}$

$\sqrt{1-4x^2}=\sqrt{1-\sin^2{t}}=\sqrt{\cos^2{t}}=|cos t|=cos t$

$\int\sqrt{1-4x^2}\,dx$
$=\frac{1}{2}\int\cos^2{t}\,dt$
$=\frac{1}{4}\int{1+\cos{2t}}\,dt$
$=\frac{1}{4}t + \frac{1}{8}\sin{2t}$

Set up triangle with angle t, adjacent $=\sqrt{1-4x^2}$, opposite=2x, and hypotenuse=1.

$\frac{1}{4}sin^{-1}{2x}+\frac{x\sqrt{1-4x^2}}{8}+C$

2. sin(2t)= 2 sin(t) cos(t). Here, you have sin(t)= 2x and, from $sin^2(t)+ cos^2(t)= 1$, $cos(t)= \sqrt{1- sin^2(t)}= \sqrt{1- 4x^2}$ so that $sin(2t)= 2(2x)(\sqrt{1- 4x^2})$.

$\frac{1}{8} sin(2t)= \frac{1}{8}\left(4x\sqrt{1- 4x^2}\right)= \frac{1}{2}x\sqrt{1- 4x^2}$.

3. Hello, Mike9182!

Your error is a silly one in the last step . . . back-substituting.

$\frac{1}{4}t + \frac{1}{8}\sin{2t} + C$

$\sin t \:=\:2x$

Set up triangle with angle $t\!:\;\;adj =\sqrt{1-4x^2},\;opp=2x,\;hyp=1$

. . This is all correct!

So we have: . $\begin{Bmatrix}\sin t \:=\:2x \\ \cos t \:=\:\sqrt{1-4x^2} \end{Bmatrix}$

Then: . $\sin 2t \:=\:2\sin t\cos t \:=\:2(2x)(\sqrt{1-4x^2}) \;=\;4x\sqrt{1-4x^2}$

Substitute: . $\frac{1}{4}\sin^{-1}(2x) + \frac{1}{8}\left(4x\sqrt{1-4x^2}\right) + C$

. . . . . . . $=\;\frac{1}{4}\sin^{-1}(2x) + \frac{1}{2}x\sqrt{1-4x^2} + C$