Evaluating $\lim\limits_{h\to 0^+} \frac1h(\int_{0}^{\pi}\sin^{h}x~{\rm d}x-\pi)$ The initial question is to find
$$\lim_{n\rightarrow \infty}n^3\left(\tan\left(\int_{0}^{\pi}\sqrt[n] {\sin x}\,{\rm d}x\right)+\sin\left(\int_{0^+}^{\pi}\sqrt[n] {\sin x}\,{\rm d}x\right)\right),$$
and I simplify it to be
$$\lim_{h\rightarrow0} \frac{1}{h}\left(\int_{0}^{\pi}\sin^{h}x\,{\rm d}x-\pi\right)=?$$
Am I wrong or any idea?
$$\lim_{n\rightarrow \infty}n^3\left(\tan\left(\int_{0}^{\pi}\sqrt[n] {\sin x}{\rm d}x\right)+\sin\left(\int_{0}^{\pi}\sqrt[n] {\sin x}{\rm d}x\right)\right)\\=
\lim_{h\rightarrow 0^+}\frac{1}{h^3}{\left(\tan\left(\int_{0}^{\pi} {\sin^h x}{\rm d}x-\pi\right)-\sin\left(\int_{0}^{\pi} {\sin^h x}{\rm d}x-\pi\right)\right)}\\=\frac{1}{2}\lim_{h\rightarrow 0^+}\frac{1}{h^3}\left({\int_{0}^{\pi}\sin^{h}x~{\rm d}x-\pi}\right)^3$$
 A: Your initial question is difficult requiring Taylor series for Gamma function and the answer appears to be $-(\pi\log 2)^3/2$.

Consider $$s_n=\int_{0}^{\pi}\sin^{1/n}x\,dx=\frac{\Gamma((n+1)/2n)\Gamma(1/2)}{\Gamma ((2n+1)/2n)}$$ and then we have $$\Gamma((n+1)/2n)=\Gamma(1/2)+\frac{1}{2n}\Gamma(1/2)\frac{\Gamma'(1/2)}{\Gamma (1/2)}+o(1/n)\\=\sqrt{\pi}-\frac{\sqrt{\pi}} {2n}(2\log2 +\gamma ) +o(1/n)$$ and $$\Gamma((2n+1)/2n)=1+\frac{1}{2n}\Gamma'(1)+o(1/n)=1-\frac{\gamma}{2n}+o(1/n)$$ and thus $$s_n=\pi\left(1-\frac{2\log 2+\gamma}{2n}+o(1/n)\right)\left(1+\frac{\gamma}{2n}+o(1/n)\right)$$ ie $$s_n=\pi+\frac{A} {n} +o(1/n)$$ where $A=-\pi\log 2$.
The desired limit is equal to the limit of $$n^3\sin(\pi-s_n)\cdot\frac{1-\cos (\pi-s_n)} {(\pi-s_n) ^2}\cdot(\pi-s_n)^2\cdot\frac{1}{\cos s_n} $$ which is same as the limit of $$-\frac{1}{2}n^3(\pi-s_n)^3$$ ie equal to $A^3/2$.

Alternatively one can work with your approach and get the desired limit as $L^3/2$ where $$L=\lim_{t\to 0}\int_{0}^{\pi}\frac{\sin^{t}x-1}{t}\,dx$$ and one can swap limits and integral sign (this needs justification) to get $$L=\int_{0}^{\pi}\log\sin x\, dx=-\pi\log 2$$
A: You can use Lebesgue Dominate convergence theorem.
Since $[0,\pi]$ is compact and we have 
$$\lim_{h\rightarrow0^+} \frac{\left(\sin^{h}x-1\right)}{h }=\frac{d}{dz}\left(\sin^z x\right)\Bigg|_{z=0} = \ln (\sin x)$$
Subsequently by mean value theorem there exists $z_h\in (0,h)$, $h>0$ such that 
  $$\frac{\left(\sin^{h}x-1\right)}{h } =  \ln (\sin x) \cdot\sin^{z_h}x$$
which is bounded for almost every $x\in (0,\pi)$ and all $h\in(0,1)$  by the integrable function $x\mapsto - \ln (\sin x)$ 
hence by Lebesgue Dominated convergence theorem we  have, 
 $$\lim_{h\rightarrow0^+} \frac{1}{h}\left(\int_{0}^{\pi}\sin^{h}x\,{\rm d}x-\pi\right)\\=\lim_{h\rightarrow0^+} \int_{0}^{\pi}\frac{\left(\sin^{h}x-1\right)}{h}\,{\rm d}x\\=\int_{0}^{\pi}\ln(\sin x)dx=-\pi\log 2$$
Indeed, 
See here Can $ \int_0^{\pi/2} \ln ( \sin(x)) \; dx$ be evaluated with "complex method"?
A: For $h>-1,$ let $I(h) = \int_0^{\pi}[(\sin x)^h - 1]\, dx.$ ($h>-1$ insures a convergent integral). I'll show 
$$\tag 1 I(h) = h\int_0^{\pi}\ln (\sin x)\, dx + O(h^2).$$
The Lagrange form of the remainder in Taylor shows
$$e^u=1+u +r(u), \,\,\text {where } |r(u)|\le e^{|u|}u^2.$$
Thus for $x\in (0,\pi),$
$$(\sin x)^h = e^{h\ln (\sin x)} = 1 + h\ln (\sin x) + r_h(x),$$
where $|r_h(x)|\le (\exp{|h\ln (\sin x)|})h^2\ln^2 (\sin x).$ We'll be done if we show $\int_0^{\pi}|r_h(x)|\,dx = O(h^2).$
Assume $|h|<1/2.$ Then
$$\exp{|h\ln (\sin x)|} = \exp{\{|h|\ln (1/\sin x)\}} = \exp{\ln (1/\sin x)^{|h|}} = \frac{1}{(\sin x)^{|h|}}\le \frac{1}{(\sin x)^{1/2}}.$$
For such $h$ this implies
$$|r_h(x)|\le h^2\frac{\ln^2 (\sin x)}{(\sin x)^{1/2}}.$$
The function of $x$ on the right is integrable over $(0,\pi),$ since the singularity at $0$ is on the order of $(\ln x)^2/x^{1/2}$ at $0;$ similarly for the singularity at $\pi.$
Putting everything together proves $(1),$ and yields the desired answers to both questions.
