Question

Evaluate: $\int \frac{1}{co{s}^{4}x+si{n}^{4}x}dx.$

Answer 1

Let $I=\int \frac{1}{co{s}^{4}x+si{n}^{4}x}dx$.

Dividing Nr and Dr by $co{s}^{4}x$, we get:

$I=\int \frac{se{c}^{4}x}{1+ta{n}^{4}x}dx=\int \frac{(1+ta{n}^{2}x)se{c}^{2}x}{1+ta{n}^{4}x}dx$ [Put $tanx=t\Rightarrow se{c}^{2}xdx=dt$

$\Rightarrow \int \frac{(1+t^2)}{1+t^4}dt\Rightarrow I=\int \frac{\frac{1}{t^2}+1}{\frac{1}{t^2}+t^2}dt$ [Dividing Nr & Dr by $t^2$

Now, put $-\frac{1}{t}+t=v\Rightarrow (\frac{1}{t^2}+1)dt=dv$. Also ${(-\frac{1}{t}+t)}^2=v^2\Rightarrow \frac{1}{t^2}+t^2=v^2+2$

$\therefore I=\int \frac{1}{v^2+(\sqrt{2}{)}^2}dv=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{v}{\sqrt{2}}\right)+C$

So, $I=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)+C\Rightarrow I=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{ta{n}^{2}x-1}{\sqrt{2}tanx}\right)+C$.