Question

integration of:
dx/(sin^4 x + cos^4 x)

Answer 1

\[\int\limits \left(\cos ^4 x+\sin ^4 x\right) \, dx\]

Answer 2

no. it is in the denominator

Answer 3

\[\int\limits \frac{1}{\cos ^4 x+\sin ^4 x} \, dx\]

Answer 4

\[\int\limits_{}^{} dx / ( \sin ^{4} + \cos ^{4} ) \]

Answer 5

try converting it to able to use u-sub...
\[\int\limits\limits \frac{1}{\cos ^4 x+\sin ^4 x} \, dx \\ \\ \Large \int\limits\limits \frac{\frac{1}{\cos^4x}}{(\cos ^4 x+\sin ^4 x )\frac{1}{\cos^4x}} \, dx\]
\[\int\limits \frac{\sec ^4 x}{\tan ^4 x+1} \, dx\]
\[\int\limits \frac{\left(1+\tan ^2 x\right) \left(\sec ^2 x\right)}{1+\tan ^4 x} \, dx \]
then let u=tanx du=sec^2x
\[\int\limits \frac{u^2+1}{u^4+1} \, du\]

Answer 6

It could also be useful to consider \[
\sin^4(x) = (1-\cos^2(x))^2
\]

Answer 7

thanks a lot. got it.