## Callisto 3 years ago Integration problem... $\int \frac{dx}{sin^4x+cos^4x}$

1. Callisto

$=\int \frac{sec^4x \ dx}{tan^4x +1}$$=\int \frac{(tan^2x+1)sec^2x \ dx}{tan^4x +1}$ $= \frac{(tan^2x+1) \ d(tanx)}{tan^4x +1}$ Let t=tanx Integral becomes$= \frac{(t^2+1) \ dt}{t^4 +1}$ How should I continue?

2. Callisto

@shivam_bhalla :(

3. blockcolder

Partial fractions, but first $$t^4+1=t^4+2t^2+1-2t^2=(t^2+1)^2-(t\sqrt{2})^2$$

4. Callisto

Why haven't my teacher taught me partial fraction....... /_\

5. 1729antony

try this divide throught out by t^2 so it becomes (1+1/t^2)/(t^2 +1/t^2).. then substitute t-(1/t) =m so the equation becomes dm/(m^2 +2) which you can integrate

6. shivam_bhalla

@1729antony , is absolutely correct. :)

7. shivam_bhalla

@blockcolder , with partial fractions, the problem becomes hectic to solve

8. blockcolder

I just remembered a problem wherein a t^4+1 came out, and I immediately thought of partial fractions, since that's what I did that time.

9. Callisto

So, here it goes~ $\int \frac{dx}{sin^4x+cos^4x} = \int\frac{sec^4x}{tan^4x+1}dx = \int \frac{tan^2x+1}{tan^4+1}d(tanx)$$= \int \frac{t^2+1}{t^4+1}dt = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$$Let \ m=t-\frac{1}{t}, \ dm = 1+\frac{1}{t^2}dt$$\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt = \int \frac{1}{m^2+2}dm = \int \frac{\sqrt2sec^2y}{(\sqrt2tany)^2+2}dy$$= \frac{\sqrt2}{2}\int \frac{sec^2y}{tan^2y+1}dy = \frac{\sqrt2}{2} \int dy = \frac{\sqrt{2}}{2}y+C$$= \frac{\sqrt{2}}{2} tan^{-1}(\frac{tanx-\frac{1}{tanx}}{\sqrt2})+C$ Sigh... Probably, it's going to be wrong :|

10. shivam_bhalla

@Callisto , you got it right. Well done :)

11. Callisto

Finally... (burst into tears...)