Question
$\text{Solve for x:} \tan^{-1} 3x + \tan^{-1} 2x = \frac{ \pi}{4}$
$\Rightarrow \frac{5x}{1 - 6x^{2}} = 1 \Rightarrow 6x^{2} + 5x - 1 = 0$
$\text{Solving to get x} = -1, \text{x} = \frac{1}{6} $
$\text{x} = -1 \text{does not satisfy the equation,} \therefore \text{x} = \frac{1}{6} \text{is the solution}$
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$\int\frac{\cos\text{x}}{\sqrt{4+\sin^2\text{x}}}\text{ dx}$
f(x) = x2, g(x) = cosx