question_answer

Evaluate \[\int{\frac{dx}{{{\sin }^{4}}x+{{\cos }^{4}}x}.}\]

Answer:

Let \[l=\int{\frac{dx}{{{\sin }^{4}}x+{{\cos }^{4}}x}}\] On dividing numerator and denominator by \[{{\cos }^{4}}x,\]we get \[l=\int{\frac{\sec }{{{\tan }^{4}}x+1}\,dx=\int{\frac{{{\sec }^{2}}x\,(1+{{\tan }^{2}}x)}{1+{{\tan }^{4}}x}}\,dx}\] \[[\because {{\sec }^{2}}x=1+{{\tan }^{2}}x]\] Put \[t=\tan x\Rightarrow dt={{\sec }^{2}}x\,dx\] \[\therefore \] \[l=\int{\frac{1+{{t}^{2}}}{1+{{t}^{4}}}\,dt=\int{\frac{1+\frac{1}{{{t}^{2}}}}{\frac{1}{{{t}^{2}}}+{{t}^{2}}}\,dt}}\] [dividing numerator and denominator by \[{{t}^{2}}\]] \[=\int{\frac{1+\frac{1}{{{t}^{2}}}}{{{\left( t-\frac{1}{t} \right)}^{2}}+2}}\,dt\] Put \[u=t-\frac{1}{t}\Rightarrow du=\left( 1+\frac{1}{{{t}^{2}}} \right)dt\] \[\therefore \]      \[l=\int{\frac{du}{{{u}^{2}}+2}}\] \[=\frac{1}{\sqrt{2}}{{\tan }^{-\,1}}\frac{u}{\sqrt{2}}+C\]      \[\left[ \because \int{\frac{dx}{{{x}^{2}}+{{a}^{2}}}=\frac{1}{a}{{\tan }^{-\,1}}\frac{x}{a}} \right]\] \[=\frac{1}{\sqrt{2}}{{\tan }^{-\,1}}\frac{\left( t-\frac{1}{t} \right)}{\sqrt{2}}+C\]            \[\left[ \because u=t-\frac{1}{t} \right]\] \[=\frac{1}{\sqrt{2}}{{\tan }^{-\,1}}\frac{({{t}^{2}}-1)}{\sqrt{2}t}\] \[=\frac{1}{\sqrt{2}}{{\tan }^{-\,1}}\left( \frac{{{\tan }^{2}}x-1}{\sqrt{2}}\tan x \right)+C\] \[[\because t=\tan x]\]