The integral _ ^08xdx4 ^2x+ ^2x is equal to

Question

The integral ${\int }_{\pi }^{0}\frac{8xdx}{4{\cos }^{2}x+{\sin }^{2}x}$ is equal to

Accepted Answer

$\begin{gathered} I={\int }_{\pi }^{0}\frac{8x}{4{\cos }^{2}x+{\sin }^{2}x}dx\ldots (1) \ I={\int }_{\pi }^{0}\frac{8(\pi -x)}{4{\cos }^{2}(\pi -x)+{\sin }^{2}(\pi -x)}dx \ I={\int }_{\pi }^{0}\frac{8(\pi -x)}{4{\cos }^{2}x+{\sin }^{2}x}dx\ldots (2) \end{gathered}$

Adding (1) and (2)

$\begin{gathered} 2I=8\pi {\int }_{\pi }^{0}\frac{1}{4{\cos }^{2}x+{\sin }^{2}x}dx \ I=4\pi 	imes 2{\int }_{\pi }^{0}\frac{{\sec }^{2}x}{4{	an }^{2}x}dx \end{gathered}$

Put $	an x=t$

${\sec }^{2}xdx=dt$

$I=8\pi {\int }_{\infty}^{0}\frac{dt}{4+{t}^{2}}$

$\begin{gathered} I=8\pi \frac{1}{2}{({	an }^{-1}\frac{t}{2})}_{\infty}^{0} \ I=4\pi (\frac{\pi }{2}) \ I=2{\pi }^{2} \end{gathered}$

Solutions · 1