$I=\int\limits_0^{\dfrac{\pi}2}\dfrac {\sin^n x}{\cos^n x}dx \hspace{3 cm} since \tan x=\dfrac {\sin x}{\cos x}$ $=\int\limits_0^{\dfrac{\pi}2}\sin^n x \cos x \text{dx}\hspace{3cm } $ since limits are also satisfied Now by beta gamma function. This can be written as $=\dfrac 12 B\Bigg(\dfrac {n+1}2,\dfrac {-n+1}2\Bigg)$ $=\dfrac 12 \dfrac {\sqrt{\dfrac {(n+1)}2}.\sqrt{\dfrac {(-n+1)}2}}{\sqrt1} ----- (1)$ Now we have another formula i.e $\dfrac 12\sqrt P\sqrt{1-P}=\dfrac 12\dfrac {\pi}{\sin p\pi} ------ (2)$ Now if $p=\dfrac {n+1}2$ $\therefore 1-p=1-\dfrac {n+1}2=\dfrac {-n}2+\dfrac12=\dfrac {(-n+1)}2$ This satisfies the LHS of eq (2) Equation (1) becomes $I=\dfrac 12\dfrac {\pi}{\sin\Bigg[\Bigg(\dfrac {n+1}2\Bigg)\pi\Bigg]}=\dfrac \pi2 \Bigg[\dfrac 1{\sin\Bigg(\dfrac{n\pi}2+\dfrac \pi2\Bigg)}\Bigg]$ $I=\dfrac \pi2\dfrac 1{\cos\dfrac {n\pi}2}\hspace {3 cm} \because\sin (90 +\theta)=\cos \theta$ $I=\dfrac \pi2 \sec\dfrac {n\pi}2$