$\int_0^{\pi / 2} \frac{1}{1+\tan ^{2020}(x)} d x=$
$\int_0^{\pi / 2} \frac{1}{1+\tan ^{2020}(x)} d x=$
$\pi$
$\frac{\pi}{2}$
$\frac{\pi}{4}$
0
Solution
$
I=\int_0^{\pi / 2} \frac{d x}{1+\tan ^{2020} x}=\int_0^{\pi / 2} \frac{\cos ^{2020} x}{\cos ^{2020} x+\sin ^{2020} x} d x
$
On applying property $\int_0^a f(x) d x=\int_0^a f(a-x) d x$, we get
$
I=\int_0^{\pi / 2} \frac{\sin ^{2020} x}{\sin ^{2020} x+\cos ^{2020} x} d x
$
On adding Eqs. (i) and (ii), we get
$
2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}
$
Hence, option (3) is correct