<s>[INST] 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!
 [/INST] Hint: $\sec^{2}x=1+\tan^{2}x$ and
$$3-2 \sin^{2}x=3-2 (1-\cos^{2}x)=3-2 \left(1-\frac{1}{\sec^{2}x}\right)=3-2 \left(1-\frac{1}{1+\tan^{2}x}\right)$$
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