Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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
Apply :
$$\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=\vert\sin x\vert=\sin x$$ whenever $x\in[0,π].$
Note thar $x^{1/2}=\pm \sqrt{x}, if x>0.$ $$I=\int_{0}^{\pi} (1-\cos^2 x)^{1/2}=\int_{0}^{\pi} \pm \sqrt{1-\cos^2 x}~dx= \int_{0}^{\pi} \pm \sin x dx=\mp \cos x|_{0}^{\pi}= \mp [-1-1]=\pm 2.$$
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: $$\int_{-\pi}^{\pi}|\sin(x)|=\int_{-\pi}^{0}-\sin(x)dx+\int_{0}^{\pi}\sin(x)dx.$$
