How to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $? I need to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $. I've checked numerically that this is true, but I haven't been able to prove it.
I've tried trigonometric substitutions. Let $\tan u= t:$
$$\int_0^1 (1+t^2)^{\frac 7 2} dt = \int_0^{\frac{\sqrt 2}{2}} (1+\tan^2 u )^{\frac 9 2} du = \int_0^{\frac{\sqrt 2}{2}} \sec^9 u \ du = \int_0^{\frac{\sqrt 2}{2}} \sec^{10} u \cos u\ du = \int_0^{\frac{\sqrt 2}{2}} \frac {\cos u}{(1-\sin^2 u)^5} du$$
Now let $\sin u = w$. Then:
$$\int_0^{\frac{\sqrt 2}{2}} \frac {\cos u}{(1-\sin^2 u)^5} du = \int_0^{\sin {\frac{\sqrt 2}{2}}} \frac {1}{(1-w^2)^5} dw.$$
This last integral is solvable using partial fraction decomposition, but even after going through all the work required I'm not really sure how to compare it with $\frac 7 2$, because of that $\sin {\frac {\sqrt{2}}{2}}$ term, which is not easy to compare.

Since $t^2 \in [0,1],\ $ we may use the Binomial expansion,
$$ \left( 1+ t^2 \right)^{7/2} = 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8 + (\text{ alternating sequence of decreasing terms with negative leading term }),$$
so,
$$ \left( 1+ t^2 \right)^{7/2} < 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8\qquad \forall\ t\in(0,1) $$
Unless I've made a calculation error, I get:
$$\ \int_0^1 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8\ dt < 3.4.$$
