4

$$I=\int \frac{dx}{\cos^3 x-\sin ^3 x}=\int \frac {(\cos x-\sin x) dx}{(\cos x -\sin x)^2 (1+\sin x \cos x)}=\int \frac{2(\cos x-\sin x)dx}{(1-\sin 2x)(2+\sin 2x)}$$ $$\implies I=\frac{2}{3} \int (\cos x - \sin x) {\left(\frac{1}{1-\sin 2x}+ \frac{1}{2+\sin 2x}\right)}dx$$ $$\implies I=\frac{2}{3} \int (\cos x - \sin x) {\left(\frac{1}{2-(\sin x+ \cos x)^2}+ \frac{1}{1+(\sin x+\cos x)^2}\right)}dx.$$ Let us use $\sin x+\cos x=t$. Finally,we get $$I=\frac{2}{3}\int\left( \frac{1}{2-t^2}+\frac{1}{1+t^2}\right)dt=\frac{1}{3\sqrt{2}} \log \left|\frac{\sqrt{2}-(\sin x+\cos x)}{\sqrt{2}+(\sin x+\cos x)}\right|+\frac{2}{3}\tan^{-1}(\sin x+\cos x)+C.$$

Looking at a solution, it looks natural and perhaps unique. The question is: How else it can be done.

Z Ahmed
  • 43,235

2 Answers2

1

One possible way is the write $$\frac 32 \frac{1}{\cos ^3(x)-\sin ^3(x)}=\frac{\cos (x)-\sin (x)}{1+(\sin (x)+\cos (x))^2}+\frac{\sec ^2\left(\frac{x}{2}\right)}{ 2- \left(\tan \left(\frac{x}{2}\right)+1\right)^2}$$

The first integral is obvious and the second is simple.

Claude Leibovici
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0

In general, with the substitution $t=\sin x+\cos x$ \begin{align} &\int\frac{1}{\cos^{2n+1}x-\sin^{2n+1}x}\ dx\\ =& \ \frac{2^n}{1+2n}\int \bigg( \frac{1}{2-t^2}+2\sum_{k=1}^n(-1)^{k}\frac{\cos a_k \cos^{n-1}2a_k}{\sec 2a_k+1-t^2} \bigg)dt\\ \end{align} with $a_k=\frac{\pi k}{2n+1}$.

Quanto
  • 97,352