Gauge Terrell

2022-07-10

Compute ${\int }_{0}^{\pi /2}\frac{\mathrm{cos}x}{2-\mathrm{sin}2x}dx$

### Answer & Explanation

Kathryn Moody

Hint:
Knowing that $\mathrm{sin}2x=2\mathrm{sin}x\mathrm{cos}x$ and ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$. The integral can be expressed as

then use substitution $x↦\frac{\pi }{2}-x$, we have

Add the two I's and let $u=\mathrm{sin}x-\mathrm{cos}x$

Cierra Castillo

$\begin{array}{rl}I& ={\int }_{0}^{\pi /2}\frac{\mathrm{cos}x}{2-\mathrm{sin}2x}dx\\ & ={\int }_{0}^{\pi /2}\frac{\mathrm{cos}x}{2-2\mathrm{sin}x\mathrm{cos}x}dx\\ & ={\int }_{0}^{\pi /2}\frac{\mathrm{cos}x}{1+{\mathrm{cos}}^{2}x-2\mathrm{sin}x\mathrm{cos}x+{\mathrm{sin}}^{2}x}dx\\ & ={\int }_{0}^{\pi /2}\frac{\mathrm{cos}x}{1+\left(\mathrm{cos}x-\mathrm{sin}x{\right)}^{2}}dx\\ & =\frac{1}{2}\left({\int }_{0}^{\pi /2}\frac{\mathrm{cos}x}{1+\left(\mathrm{cos}x-\mathrm{sin}x{\right)}^{2}}dx+{\int }_{0}^{\pi /2}\frac{\mathrm{sin}x}{1+\left(\mathrm{cos}x-\mathrm{sin}x{\right)}^{2}}dx\right)\\ & =\frac{1}{2}{\int }_{0}^{\pi /2}\frac{\mathrm{cos}x+\mathrm{sin}x}{1+\left(\mathrm{cos}x-\mathrm{sin}x{\right)}^{2}}dx\\ & =-\frac{1}{2}{\int }_{0}^{\pi /2}\frac{d\left(\mathrm{cos}x-\mathrm{sin}x\right)}{1+\left(\mathrm{cos}x-\mathrm{sin}x{\right)}^{2}}\\ & =-\frac{1}{2}\mathrm{arctan}\left(\mathrm{cos}x-\mathrm{sin}x\right){|}_{0}^{\frac{\pi }{2}}\\ & =\frac{\pi }{4}\end{array}$