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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

abruzzi26
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2 Answers2

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

Samrat Mukhopadhyay
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  • Thank you for your response. So, basically, the only place I went wrong was my choice of upper and lower limits, right? Sorry for getting back to you that late – abruzzi26 Nov 17 '15 at 07:34
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    I would like to add that using $\int \frac{\sin x}{|\cos x|}$ might be confusing; instead you can make life simpler by breaking the limits into two intervals and then choosing proper signs for $\cos x$ – Samrat Mukhopadhyay Nov 17 '15 at 15:54
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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

Alan
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  • Thank you for your response But the cos(x) in the the integral is under an absolute sign. Doesn't that eliminate any distortions by the negative?
    Sorry for getting back to you that late.     – abruzzi26 Nov 17 '15 at 07:33
  • no, it changes the function to -cos(x) where cos(x) would be negative, so when you do derivatives/antiderivatives, you have to take the minus sign into account – Alan Nov 17 '15 at 07:47