Problem with integrating $\int_0^{\pi/2}\frac{\cos^6x}{\cos^6x+\sin^6x}dx$

Question

Someone told me there is an equation $$\int_0^{\pi/2}f(sinx)dx=\int_0^{\pi/2}f(cosx)dx$$ With this equation, it's easy to get the answer$\frac{\pi}{4}$. What I want to know is why we have this equation, and if there is an alternative way to integrate it?

Answer 1

That's because $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$.

Answer 2

Because after using substitution $x=\frac{\pi}{2}-t$ we obtain: $$\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^6x-\sin^6x}{\cos^6x+\sin^6x}dx=\int\limits_{0}^{\frac{\pi}{4}}\frac{\cos^6x-\sin^6x}{\cos^6x+\sin^6x}dx+\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos^6x-\sin^6x}{\cos^6x+\sin^6x}dx=$$ $$=\int\limits_{0}^{\frac{\pi}{4}}\frac{\cos^6x-\sin^6x}{\cos^6x+\sin^6x}dx+\int\limits_{\frac{\pi}{4}}^{0}\frac{\sin^6t-\cos^6t}{\sin^6t+\cos^6t}(-dt)=0.$$

Answer 3

We have a rule of definite integral

$$\int_0^a f(x) dx = \int _ 0^a f(a-x) dx$$

So by this rule

$\int_0^{\pi/2} f(\sin x) dx = \int_0^{\pi/2} f(\sin ({\pi/2}-x)) dx = \int _ 0^{\pi/2} f(\cos x) dx$

Now come to the answer of your question

$I=\int_0^{\pi/2}\frac{\cos^6x}{\cos^6x+\sin^6x}dx=\int_0^{\pi/2}\frac{\cos^6({\frac{\pi}{2}}-x)}{\cos^6({\frac{\pi}{2}}-x)+\sin^6({\frac{\pi}{2}}-x)x}dx =\int_0^{\pi/2}\frac{\sin^6x}{\sin^6x+\cos^6x}dx$

Then

$2I=I+I=\int_0^{\pi/2}\frac{\cos^6x}{\cos^6x+\sin^6x}dx+\int_0^{\pi/2}\frac{\sin^6x}{\sin^6x+\cos^6x}dx=\int_0^{\pi/2}\frac{\sin^6x+\cos^6x}{\sin^6x+\cos^6x}dx=\frac{\pi}{2}$

$\implies 2I=\frac{\pi}{2}\implies I= \frac{\pi}{4}$

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This is the most easiest way to get the value of the integral. So I don't think you have to go any other methods to solve this kind of integral.