Trigonometric Integral of The Year

$$ In general, the integral can be solved by using the fact $\tan\left(\frac{\pi}{2}-x\right)=\cot x$ and $$ $\int_a^b f(x)\;dx=\int_a^b f(a+b-x)\;dx.$ $$ Let $$ $I=\int_0^\frac{\pi}{2} \frac{dx}{1+\tan^n x},$ $$ then $$ $\begin{aligned} I&=\int_0^\frac{\pi}{2} \frac{dx}{1+\tan^n\left(\frac{\pi}{2}+0-x\right)}\\ &=\int_0^\frac{\pi}{2} \frac{dx}{1+\cot^nx}\\ &=\int_0^\frac{\pi}{2} \frac{dx}{1+\frac{1}{\tan^nx}}\\ &=\int_0^\frac{\pi}{2} \frac{\tan^nx}{1+\tan^nx}dx. \end{aligned}$ $$ Adding the two $I$ 's yields $$ $\begin{aligned} 2I&=\int_0^\frac{\pi}{2} \frac{1}{1+\tan^n x}dx+\int_0^\frac{\pi}{2} \frac{\tan^nx} {1+\tan^nx}dx\\ &=\int_0^\frac{\pi}{2}\;dx\\ &=\frac{\pi}{2}\\ I&=\frac{\pi}{4}. \end{aligned}$ $$ Therefore $$ $\int_0^\frac{\pi}{2} \frac{dx}{1+\tan^{2014} x}=\frac{\pi}{4}.$ $$ Thus, $p+q=4+0=\boxed{4}$ . $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

easy one ... first convert tan into sin cos take cos to numerator then apply property f(a+b-x) a= 0 and b=pi/2 ...then add two equations, all the sin and cos terms will cancel and then ultimately put the limits it will come pi/4 hence ans is 4... sorry for not giving a clear solution !