Question

$\int \frac{\text{dx}}{{\text{sin}}^4\text{x}+{\text{cos}}^4\text{x}}$ is equal to

• A. $\frac{1}{\sqrt{2}}{\text{tan}}^{-1}\left(\frac{1}{\sqrt{2}}\text{tan}2x\right)+\text{C}$
• B. $\sqrt{2}{\text{tan}}^{-1}\left(\frac{1}{\sqrt{2}}\text{tan}2x\right)+\text{C}$
• C. $\frac{1}{\sqrt{2}}{\text{tan}}^{-1}\left(\frac{1}{\sqrt{2}}\text{cot}2\text{x}\right)+\text{C}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}{\text{tan}}^{-1}\left(\frac{1}{\sqrt{2}}\text{tan}2x\right)+\text{C}$

$\int \frac{\text{dx}}{{\text{sin}}^4\text{x}+{\text{cos}}^4\text{x}}=\int \frac{\text{dx}}{({\text{sin}}^2\text{x}+{\text{cos}}^2\text{x}{)}^2-2{\text{sin}}^2\text{x}{\text{cos}}^2\text{x}}=\int \frac{2\text{dx}}{2-{\text{sin}}^{2}2\text{x}}=\int \frac{2{\text{sec}}^{2}2\text{x}}{2{\text{sec}}^{2}2\text{x}-{\text{tan}}^{2}2\text{x}}\text{dx}=2\int \frac{{\text{sec}}^{2}2\text{x}}{2+{\text{tan}}^{2}2\text{x}}\text{dx}=\int \frac{\text{dt}}{(\sqrt{2}{)}^2+{\text{t}}^2}[putting\text{tan}2\text{x}=\text{t}\Rightarrow 2{\text{sec}}^{2}2\text{x}\text{dx}=\text{dt}]=\frac{1}{\sqrt{2}}{\text{tan}}^{-1}\left(\frac{\text{t}}{\sqrt{2}}\right)+\text{C}=\frac{1}{\sqrt{2}}{\text{tan}}^{-1}\left(\frac{1}{\sqrt{2}}\text{tan}2\text{x}\right)+\text{C}$