Use the substitution $u=\frac{1}{\sqrt{3}}\tan x$ to find $\int\frac{dx}{3-2\sin^2x}$ Question:
Use the substitution $u=\frac{1}{\sqrt3}\tan x$ to find $$\int\frac{dx}{3-2\sin^2x}$$
My Working:
Let $u=\frac{1}{\sqrt3}\tan x$, then
\begin{align}
\frac{du}{dx}&=\frac{1}{\sqrt3}\cdot\sec^2x+0=\frac{\sec^2x}{\sqrt3}\\
du&=\frac{\sec^2x}{\sqrt3}\cdot dx
\end{align}
Unfortunately, after that, I do not know how to proceed. Could anyone please help? Thank you!
 A: Hint: $\sec^{2}x=1+\tan^{2}x$ and
$$3-2 \sin^{2}x=3-2 (1-\cos^{2}x)=3-2 \left(1-\frac{1}{\sec^{2}x}\right)=3-2 \left(1-\frac{1}{1+\tan^{2}x}\right)$$
A: Cheese Cake's substitution made me find a fantasy way to evaluate this integral...
$$\int\frac{dx}{3-2\sin^2x}=\int\frac{dx}{3\cos^2x+\sin^2x}=\int\frac{\sec^2x}{3+\tan^2x}dx=\frac{1}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}\tan x\right)+c$$
A: Your substitution $u=\frac{1}{\sqrt 3}\tan x$ works well as below:
$$
\begin{aligned}
\int \frac{dx}{3-2 \sin ^2 x}  =& \int \frac{\frac{\sqrt{3} d u}{3 u^2+1}}{3-2\left(\frac{3 u^2}{3 u^2+1}\right)} \\
=& \sqrt{3} \int \frac{d u}{3 u^2+3} \\
=& \frac{1}{\sqrt{3}} \tan ^{-1} u+C \\
=&\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan x}{\sqrt{3}}\right)+C
\end{aligned}
$$
Alternatively, multiplying both denominator and numerator by $\sec^2x$ gives
$$
\begin{aligned}
\int \frac{dx}{3-2 \sin ^2 x} &=\int \frac{\sec ^2 x}{3 \sec ^2 x-2 \tan ^2 x} d x \\
&=\int \frac{d(\tan x)}{\tan ^2 x+3} \\
&=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan x}{\sqrt{3}}\right)+C
\end{aligned}
$$
A: Hint: $2\sin^2x=1-\cos2x$,
$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x},$$
then use substitution!
