Question

If $\int (\sqrt{\csc x + 1})dx = k \tan^{-1}(f(x)) + c$, then $\frac{1}{k} f(\frac{\pi}{6}) =$

Answer 1

-1/2

Answer 2

The integral is $I = \int \sqrt{\csc x + 1} \, dx$.

We can rewrite the integrand as $\sqrt{\frac{1}{\sin x} + 1} = \sqrt{\frac{1 + \sin x}{\sin x}}$.

We use the half-angle formulas $1 + \sin x = (\cos(x/2) + \sin(x/2))^2$ and $\sin x = 2 \sin(x/2) \cos(x/2)$.

Assuming $x$ is in a domain where $\sin x > 0$, we have $\sqrt{1+\sin x} = |\cos(x/2) + \sin(x/2)|$.

If we consider the interval $(0, \pi)$, then $x/2 \in (0, \pi/2)$, so $\cos(x/2) > 0$ and $\sin(x/2) > 0$.

Thus, $\sqrt{1+\sin x} = \cos(x/2) + \sin(x/2)$.

The integrand becomes $\sqrt{\frac{(\cos(x/2) + \sin(x/2))^2}{2 \sin(x/2) \cos(x/2)}} = \frac{\cos(x/2) + \sin(x/2)}{\sqrt{2 \sin(x/2) \cos(x/2)}}$.

We can write this as $\frac{1}{\sqrt{2}} \left( \sqrt{\frac{\cos(x/2)}{\sin(x/2)}} + \sqrt{\frac{\sin(x/2)}{\cos(x/2)}} \right) = \frac{1}{\sqrt{2}} (\sqrt{\cot(x/2)} + \sqrt{\tan(x/2)})$.

Let $v = \sqrt{\tan(x/2)}$. Then $v^2 = \tan(x/2)$.

Differentiating with respect to $x$: $2v \frac{dv}{dx} = \frac{1}{2} \sec^2(x/2) = \frac{1}{2} (1 + \tan^2(x/2)) = \frac{1}{2} (1 + v^4)$.

So $dx = \frac{4v}{1+v^4} dv$.

The integrand is $\frac{1}{\sqrt{2}} (\frac{1}{v} + v)$.

The integral becomes $I = \int \frac{1}{\sqrt{2}} (\frac{1}{v} + v) \frac{4v}{1+v^4} dv = \frac{4}{\sqrt{2}} \int \frac{1+v^2}{1+v^4} dv = 2\sqrt{2} \int \frac{1+v^2}{1+v^4} dv$.

To evaluate $\int \frac{1+v^2}{1+v^4} dv$, divide the numerator and denominator by $v^2$: $\int \frac{1/v^2 + 1}{v^2 + 1/v^2} dv$.

Let $w = v - 1/v$. Then $dw = (1 + 1/v^2) dv$.

Also, $w^2 = (v - 1/v)^2 = v^2 - 2 + 1/v^2$, so $v^2 + 1/v^2 = w^2 + 2$.

The integral becomes $\int \frac{dw}{w^2 + 2} = \int \frac{dw}{w^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{w}{\sqrt{2}}\right) + C$.

Substitute back $w = v - 1/v = \sqrt{\tan(x/2)} - \frac{1}{\sqrt{\tan(x/2)}} = \frac{\tan(x/2) - 1}{\sqrt{\tan(x/2)}}$.

So the integral is $I = 2\sqrt{2} \left( \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\frac{\tan(x/2) - 1}{\sqrt{\tan(x/2)}}}{\sqrt{2}}\right) \right) + C = 2 \tan^{-1}\left(\frac{\tan(x/2) - 1}{\sqrt{2 \tan(x/2)}}\right) + C$.

This is in the form $k \tan^{-1}(f(x)) + c$, where $k=2$ and $f(x) = \frac{\tan(x/2) - 1}{\sqrt{2 \tan(x/2)}}$.

We need to find $\frac{1}{k} f(\frac{\pi}{6})$.

$\frac{1}{k} = \frac{1}{2}$.

$f(\frac{\pi}{6}) = \frac{\tan((\pi/6)/2) - 1}{\sqrt{2 \tan((\pi/6)/2)}} = \frac{\tan(\pi/12) - 1}{\sqrt{2 \tan(\pi/12)}}$.

We calculate $\tan(\pi/12) = \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.

$\tan(\pi/12) = \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

Substitute $\tan(\pi/12) = 2 - \sqrt{3}$ into $f(\frac{\pi}{6})$:

$f(\frac{\pi}{6}) = \frac{(2 - \sqrt{3}) - 1}{\sqrt{2 (2 - \sqrt{3})}} = \frac{1 - \sqrt{3}}{\sqrt{4 - 2\sqrt{3}}}$.

We recognize the denominator $\sqrt{4 - 2\sqrt{3}} = \sqrt{(\sqrt{3})^2 - 2\sqrt{3} \cdot 1 + 1^2} = \sqrt{(\sqrt{3} - 1)^2} = |\sqrt{3} - 1|$.

Since $\sqrt{3} > 1$, $|\sqrt{3} - 1| = \sqrt{3} - 1$.

So, $f(\frac{\pi}{6}) = \frac{1 - \sqrt{3}}{\sqrt{3} - 1} = -1$.

Finally, we calculate $\frac{1}{k} f(\frac{\pi}{6}) = \frac{1}{2} \times (-1) = -\frac{1}{2}$.