Trigonometric Integrals....

They do not have palatable integral.

I beg to differ

\sin^2 x + \cos^4 x = 1 - \sin^2 x \cos^2 x

Now

\sin^2 x \cos^2 x = \left(\frac{t^2-1}{2} \right)^2

where t = \sin x - \cos x

Hence the given integral becomes

\int \frac{1}{1-\left(\frac{t^2-1}{2} \right)^2 } \ dt = \int \frac{1}{\left(\frac{3}{2}-t^2\right)\left(\frac{1}{2}+t^2\right)} \ dt

This is easily handled

Thanks sir.(very easy.)

for (1)

http://www.goiit.com/posts/list/integral-calculus-integration-of-sinx-cosx-sin-2x-cos-4x-1119365.htm

for (2)

http://www.goiit.com/posts/list/integral-calculus-please-give-me-the-answer-of-this-integral-1118958.htm#1529259