Prove that $ \int_0^1 x(\sec x)^2 \,dx < 1.$
I tried a few combinations but nothing seems to work, as the integral is quite close to 1 itself. And question strictly inhibits use of calculators to calculate ln(cos1) or tan1
The integral is tan1+ lncos1 but there is no way in my mind to get ahead with just this, so any ideas or hints will be a lot appreciated.
As mentioned in the problem statement, $$\int_0^1 x \sec^2 x \,dx = \tan 1 + \log \cos 1.$$
Expanding in a power series about $x = 0$ gives that $$\log \cos x \leq -\frac12 x^2 - \frac1{12} x^4 ,$$ so $$\log \cos 1 \leq - \frac7{12} .$$
To bound $\tan 1$ from above, we exploit the fact that $1 \approx \frac{\pi}{3}$.
Rearranging the tangent difference identity gives $$\tan 1 = \frac{\tan \frac\pi3 - \tan \left(\frac\pi 3 - 1\right)}{\tan \frac\pi3 \tan \left(\frac\pi 3 - 1\right)+ 1} = \frac{\sqrt 3 - \tan \left(\frac\pi 3 - 1\right)}{\sqrt 3 \tan \left(\frac\pi 3 - 1\right)+ 1} .$$ Since $3 + \frac18 < \pi$, we have $\frac1{24} < \frac\pi3 - 1$, and since $\tan x > x$ on $\left(0, \frac\pi2\right)$, $$\tan 1 < \frac{\sqrt 3 - \frac1{24}}{\sqrt 3 \cdot \frac1{24} + 1} = \frac{577 \sqrt 3 - 96}{573}.$$ We've now shown that $$\tan 1 + \log \cos 1 < -\frac7{12} + \frac{577 \sqrt 3 - 96}{573} ,$$ and it's routine arithmetic to verify that the right-hand side is smaller than $1$.