Evaluation of integral from textbook

Question

Integral in question

$$ \int\frac{dx}{\sqrt{\cos(x)}\sin(x)}$$

(If it helps, the original question in my textbook is to find the definite integral corresponding to this antiderivative with the limits being $0$ to $\frac{\pi}{2}$)*

Things I've tried

1. Multiplying and dividing by $\sin(x)$, (and using the Pythagorean identity,) and reaching at this integral: $$ \int{\frac{-du}{\sqrt{u}\cdot(1-u^2)}}$$ where $u = \cos(x)$. No idea how to proceed ahead; trying to use partial fractions doesn't exactly work, at least based on my knowledge of it.
2. Going by the standard approach: to write the sines and cosines in terms of tangents using the identities $$ \sin(x) = \frac{2\tan(x/2)}{1+\tan^2(x/2)}, \cos(x) = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$$ But this doesn't yield anything helpful either.

The question(s)

1. Any hints on how to proceed?
2. Can the approach 1 be dealt with using partial fractions?

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*The question is actually the following definite integral: $$ \int_{-\pi/2}^{\pi/2}{\frac{dx}{\sqrt{\cos(x)\sin^2(x)}}} \bigg(= \int_{-\pi/2}^{\pi/2}{\frac{dx}{\sqrt{\cos(x)}\cdot|\sin(x)|}}\ = 2\int_{0}^{\pi/2}{\frac{dx}{\sqrt{\cos(x)}\cdot\sin(x)}}\bigg)$$

Answer 1

Rewrite the integral as

$$ I=\int \frac{\sin x}{\sqrt{\cos x} \sin ^{2} x} d x $$

Let $y=\sqrt{\cos x}$, then $y^{2}=\cos x \Rightarrow 2 y d y=-\sin x d x$

Resolving into partial fractions yields $$ \begin{aligned} \int \frac{d y}{1-y^{4}} &=\int \frac{1}{2 }\left(\frac{1}{1+y^{2}}+\frac{1}{1-y^{2}}\right) d y \\ &=\int\left[\frac{1}{2\left(1+y^{2}\right)}+\frac{1}{4}\left(\frac{1}{1-y}+\frac{1}{1+y}\right)\right] d y \\ &=\frac{1}{2} \tan ^{-1} y+\frac{1}{4} \ln \left|\frac{1+y}{1-y}\right|+c \\ \therefore I&=-\tan ^{-1}(\sqrt{\cos x})+\frac{1}{2} \ln \left|\frac{1-\sqrt{\cos x}}{1+\sqrt{\cos x}}\right|+C \end{aligned} $$

Answer 2

Further rewrite the integral as

$$\int \frac{4\:dx}{2\sqrt{\cos x}}\cdot\frac{\sin x}{\sin^2 x} = \int_{0}^{\frac{\pi}{2}}d(\sqrt{\cos x})\cdot\frac{-4}{1-(\sqrt{\cos x})^4}$$

$$= -2\int d(\sqrt{\cos x})\cdot\left[\frac{1}{1-(\sqrt{\cos x})^2}+\frac{1}{1+(\sqrt{\cos x})^2}\right]$$

$$=-2\Bigr[\tanh^{-1}\sqrt{\cos x}+\tan^{-1}\sqrt{\cos x}\Bigr] + C$$

From here we can see that the integral diverges, which we could have also noted from the original integral having a $\sim \frac{1}{x}$ singularity near $0$.