What is the value of this definite integral? $$\int_{0}^{\pi/2}\frac{1}{1+\tan^{101}x}dx$$
I am trying to think it in this way- as x approaches $\pi /2$ tan x approaches infinity so the denominator of the function will approach 0. But summing it over the interval is not very clear to me. Next I tried to break the tan function int o sin and cos and got $\dfrac{\cos^{101}(x)}{\cos^{101}(x)+\sin^{101}(x)}$ but I am unable to move further .So what should be done in the problem?
Let $u=\frac\pi2-x\Rightarrow\mathrm{d}u=-\mathrm{d}x$.
Then, $$\begin{align}\int_0^\frac\pi2\frac1{1+\tan^{101}x}\,\mathrm{d}x&=-\int_\frac\pi2^0\frac1{1+\tan^{101}\left(\frac\pi2-u\right)}\,\mathrm{d}u\\&=\int_0^\frac\pi2\frac1{1+\cot^{101}u}\,\mathrm{d}u\\&=\int_0^\frac\pi2\frac{\tan^{101}u}{1+\tan^{101}u}\,\mathrm{d}u\\&=\int_0^\frac\pi2\frac{\tan^{101}x}{1+\tan^{101}x}\,\mathrm{d}x\\&=\int_0^\frac\pi2\mathrm{d}x-\int_0^\frac\pi2\frac1{1+\tan^{101}x}\,\mathrm{d}x\end{align}$$ From this we can now evaluate the required integral. $$\begin{align}\int_0^\frac\pi2\frac1{1+\tan^{101}x}\,\mathrm{d}x&=\frac12\int_0^\frac\pi2\mathrm{d}x\\&=\frac12\left[x\right]_0^\frac\pi2\\&=\boxed{\frac\pi4}\end{align}$$
$$I=\int_{0}^{\pi/2}\frac{1}{1+\tan^{101}x}dx$$
$$I=\int_0^{\frac{\pi}{2}}\frac{\cos^{101}(x)}{\cos^{101}(x)+\sin^{101}(x)}\,dx$$
Use the King's Property;
$$I=\int_0^{\frac{\pi}{2}}\frac{\sin^{101}(x)}{\cos^{101}(x)+\sin^{101}(x)}\,dx$$
Add both integrals;
$$2I=\int_0^{\frac{\pi}{2}}\frac{\sin^{101}(x)+\cos^{101}(x)}{\cos^{101}(x)+\sin^{101}(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}$$
$$I=\frac{\pi}{4}$$
The original integral can be generalized (similar steps);
$$I=\int_{0}^{\pi/2}\frac{1}{1+\tan^{n}x}dx=\frac{\pi}{4}$$
