Improper integrals - where am I going wrong?

Question

The question asks:
$\int_0^\pi \frac{sinx}{\sqrt{|cosx|}}$

My attempt:
$\lim\limits_{t \to \frac{\pi}{2}^-} \int_0^t \frac{sinx}{\sqrt{|cosx|}}$ + $\lim\limits_{t \to \frac{\pi}{2}^+} \int_t^\pi \frac{sinx}{\sqrt{|cosx|}}$

$\int \frac{sinx}{\sqrt{|cosx|}} = -2\sqrt{|cosx|}$

$\lim\limits_{t \to \frac{\pi}{2}^-} [-2\sqrt{|cosx|}]_0^t = 2$
$\lim\limits_{t \to \frac{\pi}{2}^+} [-2\sqrt{|cosx|}]_t^\pi = -2$
$\lim\limits_{t \to \frac{\pi}{2}^-} [-2\sqrt{|cosx|}]_0^t + \lim\limits_{t \to \frac{\pi}{2}^+} [-2\sqrt{|cosx|}]_t^\pi = 0$

My final answer is $0$, but the marking scheme shows the final answer as $4$.

I also notice the following step from the marking scheme:
$\int_0^\pi \frac{sinx}{\sqrt{|cosx|}}$ = $2\int_0^\frac{\pi}{2} \frac{sinx}{\sqrt{|cosx|}}$
I do not understand how the marking scheme arrived to this step.

I have been struggling with this question for too long now. Could someone please help me get to the correct answer? I will appreciate that.
Thanks

Answer 1

Note that after $\frac {\pi} 2$, cosine is negative, so the absolute value function in your second integral will have a -cos(x) in the bottom. That'll add an extra - to your antiderivative of the second part, which flips it from -2 to 2

Answer 2

Hint: Note that $$\int_{0}^{\pi/2}\frac{\sin x}{\sqrt{|\cos x|}}dx=\int_{0}^{\pi/2}\frac{\sin x}{\sqrt{\cos x}}dx=2\\ \int_{\pi/2}^{\pi}\frac{\sin x}{\sqrt{|\cos x|}}dx=\int_{\pi/2}^{\pi}\frac{\sin x}{\sqrt{-\cos x}}dx=\int_{0}^{\pi/2}\frac{\cos x}{\sqrt{\sin x}}dx=2$$