Integrate the following function:

Question

Evaluate: $$\int \frac{1}{ \cos^4x+ \sin^4x}dx$$

Tried making numerator $\sin^2x+\cos^2x$
making numerator $(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$
Dividing throughout by $cos^4x$

Thank you in advance

Answer 1

Another way:

$$I=\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{(1+\tan^2x)\sec^2x dx}{1+\tan^4x}$$

Setting $\displaystyle\tan x=u,$

$$I=\frac{(1+u^2)du}{1+u^4}=\int\frac{1+\dfrac1{u^2}}{u^2+\dfrac1{u^2}}du$$

$$=\int\frac{1+\dfrac1{u^2}}{\left(u-\dfrac1u\right)^2+2}du$$

Set $\displaystyle u-\dfrac1u=v$

Answer 2

Let me try something mixing the work of chndn and lab bhattacharjee. First let's rewrite the denominator as chndn has tried.

$$\int\dfrac1{\cos^4x+\sin^4x}dx=\int\dfrac1{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x}dx=$$ $$\int\dfrac1{1-\frac12\sin^22x}dx$$

Now let's do a move similar to lab bhattacharjee's.

$$\int\dfrac{\csc^22xdx}{\csc^22x-\frac12}=\int\dfrac{\csc^22xdx}{\cot^22x+\frac12}$$

Now let $u=\cot 2x$.