Question

integration of sin 2x /sin⁴x+cos⁴x dx

Answer 1

Hello!

$\int \dfrac{\sf sin(2x)}{\sf sin^{4}x + \sf cos^{4}x} \sf dx$

⇒ $\int \dfrac{\sf sin(2x)}{(\sf sin^{2}x)^{2} + \sf cos^{4}x} \sf dx$

⇒ $\int \dfrac{\sf sin(2x)}{ (\sf 1 - cos^{2}x)^{2} + \sf cos^{4}x } \sf dx$

⇒ $\int \dfrac{\sf sin(2x)}{(\sf 1 - 2cos^{2}x + \sf cos^{4}x + \sf cos^{4}x) } \sf dx$

⇒$\int \dfrac{\sf sin(2x)}{ (\sf 2cos^{4}x - \sf 2cos^{2}x + \sf 1)} \sf dx$

⇒$\int \dfrac{\sf sin(2x)}{ (\sf 4cos^{4}x - \sf 4cos^{2}x + \sf 1) + \sf 1 / 2} \sf dx$

⇒$\int \dfrac {\sf 2sin(2x)}{ (\sf 2cos^{2}x - 1)^{2} + \sf 1} \sf dx$

⇒$\int \dfrac{\sf 1}{(\sf 2cos^{2}x - \sf 1)^{2} + \sf 1} \sf 4\: \sf cosx \: \sf sinx \: \sf dx$

Now,

$\sf 2cos^{2}x - 1$ = $\sf u$

⇒ $\sf - 4\:cosx\: sinx\: dx$ = $\sf du$

Then,

$\int \dfrac {\sf 1}{(\sf 2cos^{2}x - \sf 1)^{2} + \sf 1} \: 4 \: \sf cosx \: \sf sinx \: \sf dx$

$\int \dfrac{\sf 1 }{ (\sf u^{2} + \sf 1)} (\sf - du)$

⇒$- \int \dfrac{\sf 1} {(\sf u^{2} + \sf 1) } \sf du$

⇒$\sf - arctan \:u + C$

⇒ $\boxed{\sf - arctan(\sf 2cos^{2}x - \sf 1) + \sf C}$

Cheers!