# Trignometry Integration Problem with $\cos^2(x)$ [closed]

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

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

Apply :

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

• @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?? Feb 29, 2020 at 17:54

Note thar $$x^{1/2}=\pm \sqrt{x}, if x>0.$$ $$I=\int_{0}^{\pi} (1-\cos^2 x)^{1/2}=\int_{0}^{\pi} \pm \sqrt{1-\cos^2 x}~dx= \int_{0}^{\pi} \pm \sin x dx=\mp \cos x|_{0}^{\pi}= \mp [-1-1]=\pm 2.$$

• This is not correct. Roots are positive (at least by convention) Feb 29, 2020 at 17:05
• Yes, you are right, I have edited it now, thanks. Feb 29, 2020 at 17:11
• It's still wrong. It's either always positive or always negative. $\sin$ oscillates. Feb 29, 2020 at 17:11
• $\sin x$ does not change sign in $[o,\pi]$. Feb 29, 2020 at 17:16
• Ah, I wasn't paying attention to the range. Fair enough. Feb 29, 2020 at 17:17

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: $$\int_{-\pi}^{\pi}|\sin(x)|=\int_{-\pi}^{0}-\sin(x)dx+\int_{0}^{\pi}\sin(x)dx.$$

• Oops, I didn't notice that he changed his answer