Integral $\int\frac{\sin^4 x + \cos^4 x}{\sin^3 x + \cos^3 x} dx$

Question

integral of $\displaystyle \int \frac{\sin^4 x + \cos^4 x}{\sin^3 x + \cos^3 x}dx$

$$\int \frac{\sin^4 x + \cos^4 x} {\sin^3 x + \cos^3 x} dx$$

OR $\displaystyle \int \frac {\sin x \cos x}{\sin^3 x + \cos^3 x}$

Apologies for not mentioning the context of the question. It was some years ago. I was trying to solve an integration problem on an educational application. I was able to reduce the original question to the integral I had asked here. Now I don't remember the context. But I think I was able to provide a solution with the answers mentioned here. Thank you all and math.stackexchange.

Answer 1

Since it is your first question I will be nice this time. Next time, please learn how to write mathjax code to make your math look better, and please show some effort or give some context. That will make your questions much better.

In both your integrals, let $$ u=\cos x-\sin x. $$ The first one will transform into $$ \int\Bigl(\frac{2/3}{1+u^2}+\frac{1/3}{u^2-2}-1\Bigr)\,du $$ while the second one will be $$ \int\frac{1-u^2}{u^4-u^2-2}\,du. $$ I'm sure you can take care of these.

Answer 2

$$a^{n+1}+b^{n+1}=(a+b)(a^n+b^n)-ab(a^{n-1}+b^{n-1})$$ implies $$\dfrac{\sin^4 x + \cos^4 x}{\sin^3 x + \cos^3 x}=(\sin x+\cos x)-\dfrac{\sin x \cos x}{\sin^3 x + \cos^3 x}$$ and $$\sin^3 x + \cos^3 x=(\sin x+\cos x)(1-\sin x\cos x).$$ Therefore $$-\dfrac{\sin x \cos x}{\sin^3 x + \cos^3 x}=\dfrac{1}{\sin x +\cos x}-\dfrac{1}{\sin^3 x + \cos^3 x}.$$

Answer 3

Since $\dfrac{\sin^4 x + \cos^4 x}{\sin^3 x + \cos^3 x}=(\sin x+\cos x)-\dfrac{\sin x \cos x}{\sin^3 x + \cos^3 x}$, we only need to evaluate $$ A:=\int \frac{\sin x \cos x}{\sin ^{3} x+\cos ^{3} x} d x. $$ First of all, we split the integrand into 2 parts. $$ \frac{3 \sin x \cos x}{\sin ^{3} x+\cos ^{3} x}=-\frac{1}{\sin x+\cos x}+\frac{\cos x+\sin x}{1-\sin x \cos x} $$ Then $$ 3 A=-\underbrace{\int \frac{d x}{\sin x+\cos x}}_{J}+\underbrace{\int \frac{\cos x+\sin x}{1-\sin x \cos x} d x}_{K} $$ $$ \begin{aligned} J &=\int \frac{d x}{\sin x+\cos x}=\frac{1}{\sqrt{2}} \int \frac{d x}{\cos \left(x-\frac{\pi}{4}\right)} \\ &=\frac{1}{\sqrt{2}} \ln \left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right)\right|+c_{1} \\&=\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2}+\sin x-\cos x}{\cos x+\sin x}\right| +c_1\\ K &=2 \int \frac{d(\sin x-\cos x)}{(\sin x-\cos x)^{2}+1} \\ &=2 \tan ^{-1}(\sin x-\cos x)+c_{2} \end{aligned} $$ Now we can conclude that $$A= \frac{1}{3\sqrt{2}} \ln \left|\frac{\sqrt{2}+\sin x-\cos x}{\cos x+\sin x}\right|+ \frac{2}{3} \tan ^{-1}(\sin x-\cos x) +C$$ Hence $\boxed{\displaystyle \int \frac{\sin^4 x + \cos^4 x}{\sin^3 x + \cos^3 x}dx=\sin x-\cos x+ \frac{1}{3\sqrt{2}} \ln \left|\frac{\sqrt{2}+\sin x-\cos x}{\cos x+\sin x}\right|+ \frac{2}{3} \tan ^{-1}(\sin x-\cos x) +C }$