Integration

cos⁵x + sin⁵x = (cosx+sinx) [ cos⁴x-cos³xsinx+cos²xsin²x-cosxsin³x+sin⁴x]

= (cosx+sinx) [ (cos²x+sin²x)² - cos²x sin²x - cosx sinx (cos²x + sin²x)]

= (cosx+sinx) (1 - cosx sinx - cos²x - sin²x)

Therefore I = ∫ (sinx + cosx)dx / (1+2sinx cosx) (1-cosx Sinx - cos²x Sin²x)

Put Sinx - cosx =t, so that
(cosx + sinx) dx = dt , and
t² = 1-2sinx cosx

I = ∫ dt / (2 - t²) [ 1 - (t²-1) / 2 - (t²-1)² / 4]

I = 4 ∫ dt / (2-t²) (5-t⁴)
Using partial fraction,

I = ∫ [ 1 / (2-t²) + 1 / (2-√5) (1/2√5) (1 / √5- t²) + (1 / 2+√5) (1/2√5) (1 / √5 + t²) ] dt

=√2 log|(√2 + t) / (√2-t) | (1 / 2-√5) (1 / 5 ^3/4 ) log | (5 ^1/4 + t) / (5 ^1/4 - t) | + (1 / 2+√5) (2 / 5 ^3/4) tan ^-1 (t / 5 ^1/4) + C

thanks...