How to integrate a special type of square root function?

Question

I am trying to integrate a special type of square root function $$ f(x) = \sqrt{R^2-x^2} $$ Where $R$ is constant.
And I want to integrate it with respect to $x$. $$ \int_{0}^{R}\sqrt{R^2-x^2} dx $$ And if I do this: $$ u = \sqrt{R^2-x^2}\\ x = \sqrt{R^2-u^2} $$ But this reverses also I don't know any other concept to solve integrals. Any help will be appreciated.
Note:This is not a homework or exercise question.

Answer 1

HINT

Since $-R \leq x \leq R$, you can make the substitution $x = R\sin(y)$. Then it results that \begin{align*} \int_{0}^{R}\sqrt{R^{2} - x^{2}}\mathrm{d}x & = \int_{0}^{\pi/2}R\sqrt{R^{2} - R^{2}\sin^{2}(y)}\cos(y)\mathrm{d}y\\\\ & = \int_{0}^{\pi/2}R^{2}\sqrt{1 - \sin^{2}(y)}\cos(y)\mathrm{d}y\\\\ & = \int_{0}^{\pi/2}R^{2}\cos^{2}(y)\mathrm{d}y\\\\ & = \frac{R^{2}}{2}\int_{0}^{\pi/2}[\cos(2y) + 1]\mathrm{d}y \end{align*}

Can you take it from here?

Answer 2

Another way to do this integral is geometrically.

Note that the integral is the area of a quarter-circle with radius $R$.

Therefore, the integral is $\frac{1}{4} \pi R^2$.