Question

Integrate sin^4 x + cos^4 x / sin^3x + cos^3 x dx

Answer 1

 This was tough.  Perhaps there is a simpler method for this..I did this as follows...
As the power of terms in the numerator is higher than those in the denominator, First find the partial fractions.  Then integrate.

$f(x)=\frac{sin^4x+cos^4x}{sin^3x+cos^3x}=\frac{(sin^2x+cos^2x)^2-2Sin^2xcos^2x}{(sinx+cosx)(sin^2x-sinx\ cosx+cos^2x)}\\\\=\frac{1-sin^22x/2}{(sinx+cosx)(1-sin2x/2)}=\frac{2-sin^22x}{(sinx+cosx)(2-sin2x)}\\\\=\frac{(2+sin2x)(2-sin2x)-2}{(sinx+cosx)(2-sin2x)}\\\\=\frac{2+sin2x}{sinx+cosx}-\frac{2}{(sinx+cosx)(2-sin2x)}\\\\f(x)= \frac{1+(sinx+cosx)^2}{(sinx+cosx)}-\frac{1}{\sqrt2sin(\frac{\pi}{4}+x)(1-sinx\ cosx)}$

let  (π/4 + x) = y
 we use the identities as below:
       cos x + sin x = √2 Sin (π/4 +x)
       cos x - sin x = √2 Sin (π/4 - x)  = √2 Cos (π/4 + x)
      (cos x  - sin x)² = 2 (1 - Sin² y)

$f(x) = \frac{1}{\sqrt2}cosec(\frac{\pi}{4}+x)+\sqrt2Sin(\frac{\pi}{4}+x)-\frac{2}{\sqrt2Sin(\frac{\pi}{4}+x)[1+(sinx-cosx)^2]}\\\\=\frac{1}{\sqrt2}cosec\ y+\sqrt2\ Sin\ y-\frac{\sqrt2}{Sin\ y [1+2Sin^2(\frac{\pi}{2}-y)]}\\\\=\frac{1}{\sqrt2}cosec\ y+\sqrt2\ Sin\ y-\frac{\sqrt2}{Sin y *(3-2Sin^2y)}\\\\=\frac{1}{\sqrt2}cosec\ y+\sqrt2\ Sin\ y-\frac{\sqrt2}{3Sin y}-\frac{2\sqrt2siny}{3(3-2Sin^2y)}\\\\f(x) = \frac{1}{3\sqrt2}cosec\ y+\sqrt2sin\ y+\frac{-2\sqrt2 siny}{3*(1+(\sqrt2Cosy)^2)}$

Now integrate the expression easily now.   The answer is :

$I=-\frac{1}{3\sqrt2}Ln|Cosec\ y+Cot\ y |-\sqrt2\ Cos\ y+\frac{2}{3}Tan^{-1}(\sqrt2*Cosy),\\\\where\ y=\frac{\pi}{4}+x\\\\\I=\frac{1}{3\sqrt2}\ Ln | tan(\frac{\pi}{8} + \frac{x}{2})| - cos x + sin x + \frac{2}{3}\ Tan^{-1}(Cos x - Sin x)$

Now you can verify this answer by differentiating and adding the terms.

I hope that is understood with not much difficulty.  It seems a bit lengthy and tough.