Question

Integrate ( sin^4x + cos^4x ) / ( sin^3x + cos^3x )

Answer 1

Integration of f(x) dx
  = ∫ (Sin⁴ x + Cos⁴ x) dx / (Sin³ x + Cos³ x) 

Let u = sin x - cos x.
      du = (sin x + cos x) dx
     u² = (sin x - cosx)² = 1 - 2 sinx  cosx
     sin x * cos x = (1 - u²)/2

$f(x)=\frac{sin^4x+cos^4x}{sin^3+cos^3x}\\\\=cosx+sinx-\frac{(cos^2x+sin^2x)sinx \: cosx}{(sinx+cosx)(sin^2x-sinx \: cosx+cos^2)}\\\\2nd \: term*dx=-\frac{sinx \: cosx}{(sinx+cosx)(1-sinx * cosx)}dx\\\\=-\frac{(1-u^2)/2}{(sinx+cosx)^2)(1-\frac{1-u^2}{2})}du\\\\=-\frac{1-u^2}{(1+1-u^2)(1+u^2)}du\\\\=\frac{1}{3} [ \frac{1}{2-u^2} -\frac{2}{1+u^2} ]du\\\\Integral \: is=\frac{1}{3*2\sqrt{2}}Ln|\frac{\sqrt{2}+u}{\sqrt{2}-u} |-\frac{2}{3}Tan^{-1}u+K$

$\\\\So\: \int {f(x) dx} \: dx= Sinx - Cosx \\\\+\frac{1}{6\sqrt{2} } Ln | \frac{\sqrt{2}+sinx-cosx}{\sqrt{2}+cosx-sinx} |\\\\-\frac{2}{3}Tan^{-1}(sinx-cosx)+K$

Answer is
  I = Sin x - Cos x + 1/6√2 * Ln | (√2+ Sinx - cosx)/(√2 - sinx + cosx) | - 2/3 * Tan⁻¹ (sinx+cosx) + K
     OR,  Sin x - Cos x + 1/3√2 * Ln | Tan(3π/8 - x/2) | -2/3 * Tan⁻¹ (sinx - cosx) + K