Question

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

Answer 1

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

Answer 2

@shivam_bhalla :(

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

@1729antony , is absolutely correct. :)

Answer 7

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

Answer 8

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.

Answer 9

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 :|

Answer 10

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

Answer 11

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