Question

Evaluate: $\int \frac{\tan x}{\sqrt{{\sin }^{4}x+{\cos }^{4}x}}dx=$

Answer 1

$\int \frac{tanx}{\sqrt{{sin}^{4}x+{cos}^{4}x}}dx$

$\Rightarrow {sin}^{4}x+{cos}^{4}x={({sin}^{2}x+{cos}^{2}x)}^2-2{sin}^{2}x\times {cos}^{2}x$

$=1-{2sin}^{2}x{cos}^{2}x$

$\int \frac{tanx}{\sqrt{1-2{sin}^{2}x{cos}^{2}x}}dx$

$sinx=\frac{tanx}{secx},cosx=\frac{1}{secx}$

$\int \frac{tanx}{\sqrt{1-\frac{{2tan}^{2}x}{{sec}^{4}x}}}$

$\Rightarrow \int \frac{{sec}^{2}xtanx}{\sqrt{{(1+{tan}^{2}x)}^2-2{tan}^{2}x}}dx$

$\Rightarrow \int \frac{{sec}^{2}xtanx}{\sqrt{1+{tan}^{4}x}}dx$

$tanx=m$

${sec}^{2}xdx=dm$

$\int \frac{m}{\sqrt{1+m^4}}dm$

Putting $m^2=t$

$2mdm=dt$

$=\frac{1}{2}\int \frac{dt}{\sqrt{1+t^2}}$

Standard integral

$=\frac{1}{2}In|\sqrt{t^2+1}+t|+c$

$\Rightarrow \frac{1}{2}In|\sqrt{m^4+1}+m^2|+c$

$\Rightarrow \frac{1}{2}In|\sqrt{{tan}^{4}x+1}+{tan}^{2}x|+c$