Question

Evaluate the integral

${\int }_{\pi }^{5\pi /4}\frac{\sin 2x}{{\cos }^{4}x+{\sin }^{4}x}dx$

• A. $\frac{5\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

${\int }_{\pi }^{\frac{3\pi }{4}}\frac{sin2x}{co{s}^{4}x+si{n}^{4}x}dx={\int }_{\pi }^{\frac{3\pi }{4}}\frac{dsi{n}^{2}x}{(1-si{n}^{2}x{)}^2+si{n}^{4}x}$

$={\int }_{\pi }^{\frac{3\pi }{4}}\frac{dsi{n}^{2}x}{(2si{n}^{2}x-2si{n}^{2}x+\frac{1}{2}+\frac{1}{2})}$

$si{n}^{2}x=t$${\int }_0^{\frac{1}{2}}\frac{dt}{(\sqrt{2}t-\frac{1}{\sqrt{2}}{)}^2+(\frac{1}{\sqrt{2}}{)}^2}=\frac{1}{\sqrt{2}}\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}ta{n}^{-1}{\left[\frac{(\sqrt{2}t-\frac{1}{\sqrt{2}})}{\frac{1}{\sqrt{2}}}\right]}_0^{\frac{1}{2}}$

$=ta{n}^{-1}(2t-1{)}_0^{1/2}$

$=ta{n}^{-1}\left(0\right)-ta{n}^{-1}(-1)$

$=0-\left(\frac{-\pi }{4}\right)$

$=\frac{\pi }{4}$