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I'm stuck on the following problem:

$$\int_0^\pi\left(1-\cos^2x\right)^{0.5}\,\mathrm{d}x.$$

The first step I took was to break down $\cos^2(x)$ into $0.5(1+\cos(2x))^{0.5}$ but I don't know what to do past this point.

A step through would be greatly appreciated

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    $\begingroup$ $$1-\cos^2x=\sin^2x$$ $\endgroup$ Commented Feb 29, 2020 at 16:59
  • $\begingroup$ Your differential $\mathrm dx$ is missing. You don't integrate a mere function -- you integrate the differential. $\endgroup$
    – Allawonder
    Commented Feb 29, 2020 at 17:19

3 Answers 3

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Apply :

$$\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=\vert\sin x\vert=\sin x$$ whenever $x\in[0,π].$

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  • $\begingroup$ @DonThousand Next time, don't answer in a comment, maybe, and post an answer, rather than trying to take credit for the answer from another user who does post an answer?? $\endgroup$
    – amWhy
    Commented Feb 29, 2020 at 17:54
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Note thar $x^{1/2}=\pm \sqrt{x}, if x>0.$ $$I=\int_{0}^{\pi} (1-\cos^2 x)^{1/2}=\int_{0}^{\pi} \pm \sqrt{1-\cos^2 x}~dx= \int_{0}^{\pi} \pm \sin x dx=\mp \cos x|_{0}^{\pi}= \mp [-1-1]=\pm 2.$$

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  • $\begingroup$ This is not correct. Roots are positive (at least by convention) $\endgroup$ Commented Feb 29, 2020 at 17:05
  • $\begingroup$ Yes, you are right, I have edited it now, thanks. $\endgroup$
    – Z Ahmed
    Commented Feb 29, 2020 at 17:11
  • $\begingroup$ It's still wrong. It's either always positive or always negative. $\sin$ oscillates. $\endgroup$ Commented Feb 29, 2020 at 17:11
  • $\begingroup$ $\sin x$ does not change sign in $[o,\pi]$. $\endgroup$
    – Z Ahmed
    Commented Feb 29, 2020 at 17:16
  • $\begingroup$ Ah, I wasn't paying attention to the range. Fair enough. $\endgroup$ Commented Feb 29, 2020 at 17:17
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Dr Ahmed's answer is good, but there's an additional detail I think you might want to heed for future problems.

Usually, when we take the square root of a squared expression, we have $\sqrt{x^2}=|x|$. So in the third step of Dr Ahmed's solution we should have $\int_0^\pi \sqrt{\sin^2(x)} dx=\int_0^\pi |\sin(x)| dx$.

Since $\sin(x)$ is positive for all $x\in[0,\pi]$, we can remove the absolute value bars and just have $$\int_0^\pi |\sin(x)|dx=\int_0^\pi \sin(x) dx=2.$$


If the integral's lower bound covered a region where $\sin(x)$ could yield negative values, then we'd need to split the integral appropriately. For instance: $$\int_{-\pi}^{\pi}|\sin(x)|=\int_{-\pi}^{0}-\sin(x)dx+\int_{0}^{\pi}\sin(x)dx.$$

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  • $\begingroup$ Oops, I didn't notice that he changed his answer $\endgroup$
    – Hadi
    Commented Feb 29, 2020 at 17:30

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