Show that $\int_0^{\frac{π}{2}}(1+\sin^2 x)\,dx <\frac{3π}{4}$

Question

Show that :

$\int_0^{\frac{π}{2}}(1+\sin^2 x)dx <\frac{3π}{4}$

Without calculus integral of

$\int_0^{\frac{π}{2}}\sin^2x\,dx$

Can $\sin x≤x$ or $\sin x≤\tan x\:$ work here !!

Answer 1

=The integral can be split into two parts, $\int_0^\frac{\pi}{2}1dx=\frac{\pi}{2}$ and $\int_0^\frac{\pi}{2}\sin^2xdx$. We know $\sin^2x+\cos^2x=1$ and $\int_0^\frac{\pi}{2}\cos^2x\,dx=\int_0^\frac{\pi}{2}\sin^2x\,dx$ Therefore $\int_0^\frac{\pi}{2}\sin^2x\,dx=\frac{\pi}{4}$. As a result $\int_0^\frac{\pi}{2}1+\sin^2x\,dx=\frac{3\pi}{4}$, not $\lt$ as proposed.

Answer 2

$sin^2(x)={{1-cos(2x)}\over 2}$ implies that $1+sin^2(x)={3\over 2}-{{cos(2x)}\over 2}$

$\int_0^{{\pi\over 2}}{{cos(2x)}\over 2}dx=0$ and $\int_0^{\pi\over 2}{3\over 2}dx={{3\pi}\over 4}$