How to evaluate $\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{2013}x}$

Question

How to evaluate the integral

$$\int_0^{\frac{\pi}{2}}\frac{\mathrm dx}{1+\tan^{2013}x}$$

It is almost impossible to calculate the antiderivative of this function by hand. Are there any tricks?

Answer 1

$$\text{Use }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$

Generalization : $$\text{If }J=\int_a^b\frac{g(x)}{g(x)+g(a+b-x)}dx, J=\int_a^b\frac{g(a+b-x)}{g(x)+g(a+b-x)}dx$$

$$\implies J+J=\int_a^b dx$$ provided $g(x)+g(a+b-x)\ne0$

Here $\displaystyle J=\int_0^{\frac\pi2}\frac{dx}{1+\tan^{2013}x}=\int_0^{\frac\pi2}\frac{\cos^{2013}x}{\sin^{2013}x+\cos^{2013}x}dx$ so $g(x)=\cos^{2013}x$

and in $\displaystyle\left[0,\frac\pi2\right]\sin x,\cos x\ge0\implies \cos^{2013}x+\sin^{2013}x>0$ i.e., $\ne0$

Answer 2

Use the identity $$\int_{0}^{\pi/2}f(x)\,\,\mathrm{d}x = \int_{0}^{\pi/2}f\left(\frac{\pi}{2} - x\right)\,\,\mathrm{d}x $$

Now define $$g(x) = f(\pi/2 - x) $$ and note that $f(x) + g(x) = 1$. Thus the integral must equal $\pi/4$, check this yourself!

Note that the exponent 2013 is a red herring - this technique works owing to the symmetry of the tangent, and 2013 could be replaced by any real number and the integral would still equal $\pi/4$.