Is this kind of inverse substitution justified?

Question

I'm trying to prove that for every integer $n\geq 0$ we have $$\int_0^{\pi/2} (1+\cos t)^ndt\geq \frac{2^{n+1}-1}{n+1}.$$ I started out by rewriting the RHS as $\int_0^{1}(1+x)^ndx$ and substituting $x=\cos t$, where $t$ ranges from $\pi/2$ to $0$. Then $dx=-\sin t dt$ and we get $$\int_{\pi/2}^0(1+\cos t)^n (-\sin t)dt = \int_0^{\pi/2}(1+\cos t)^n \sin t dt \leq \int_0^{\pi/2}(1+\cos t)^n dt,$$ where in the equality I took out the minus sign from the $\sin$ to flip the limits of integration. What seems suspicious to me though is how $t$ ranges from the larger value $\pi/2$ to $0$: is this ok? I tried doing different substitutions like $x=\cos t$ but for $-\pi/2\leq t\leq 0$ but I couldn't recover what I wanted from these.

Answer 1

Hint

Put $$x=1+\cos(t)$$ with $$dx=-\sin(t)$$

Your integral satisfies

$$\int_0^{\frac{\pi}{2}}(1+\cos(t))^ndt\ge \int_0^\frac{\pi}{2}(1+\cos(t))^n\sin(t)dt$$

because $$(\forall t\in[0,\frac{\pi}{2}])\;\;\; 0\le\sin(t)\le 1$$

With the substitution above, the right integral becomes

$$-\int_2^1x^ndx=\int_1^2x^ndx=\Bigl[\frac{x^{n+1}}{n+1}\Bigr]_1^2$$