Proving that the series with the general term $u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$ diverges. 
$$u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$$ for $ n \in \mathbb{N}^*$.
  
  
*
  
*Prove that $(u_n)$ is convergent toward $0$. 
  
*Prove that the series with the general term $(-1)^n u_n$ converges. 
  
*Prove that $$\sum_{n=0}^{\infty} (-1)^n u_n = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx$$ 
  
*Compute $$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx$$ 
Hint: You can start by proving $$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos(x)}dx$$ 
  
*Prove that the series with the general term $u_n$ diverges. 
Hint: You can start by proving $$u_n \geq \frac{1}{n + 1}$$




*

*For $ 0 \leq x \leq \frac{\pi}{2}$ we have:  $0 \leq u_n \leq \frac{\pi}{2}$ wich I can't derive the convergence from it. How can I prove it is convergent toward 0? 


I am stuck with question 4. and 5, I could not see how to prove the hints.
 A: To prove that $u_n$ converges to $0$, you can split the integral into two parts, one near $\pi/2$ (where $\sin$ is close to $1$), and the rest. Namely, take any $\delta >0$ small, and write
$$
\int_0^{\pi/2} \sin^n x dx = \int_0^{\frac{\pi}{2} - \delta} + \int_{\frac{\pi}{2} - \delta}^{\pi/2} : = I_1 + I_2. \tag{1}
$$
Since $|\sin(x) | \leq 1$ we have $|I_2| \leq \delta$. In view of monotonicity of $\sin x$ in $[0,\pi/2]$ we get
$$
|I_1| \leq \frac{\pi}{2} \left(\sin {(\frac{\pi}{2} - \delta)}\right)^n.\tag{2}
$$ 
But since $\delta> 0$ is fixed and $0<\sin {(\frac{\pi}{2} - \delta)} < 1$ from $(2)$ we get that $I_1 \to 0$ as $n\to \infty$. Putting this and the estimate of $I_1$ with arbitrary smallness of $\delta>0$ implies that $u_n \to 0$.
As for divergence of the series, you can use the well-known inequality $\frac{2}{\pi}x < \sin x$ where $(0,\pi/2)$ from which we get that 
$$
u_n \geq \left(\frac{2}{\pi}\right)^n \int_0^{\pi/2} x^n dx = \left(\frac{2}{\pi}\right)^n \frac{1}{n+1} \left(\frac{\pi}{2}\right)^{n+1} = \frac{\pi}{2} \frac{1}{n+1}.
$$
Thus $u_n$ is bounded below by harmonic series which is divergent, hence so is $\sum_n u_n$.
A: Using integration by parts it is easy to prove that $u_n=((n-1)u_{n-2})/n$. Using this relation it follows that $u_n$ is decreasing and clearly it is non-negative so that $u_n$ is convergent. Using this recurrence relation we will prove that sequence $a_{n} =u_{2n}$ tends to $0$. We have $$\frac{a_{n}} {a_{n-1}}=\frac{2n-1}{2n}$$ and hence on taking logs we have $$\log a_n-\log a_{n-1}=\log\left(1-\frac{1}{2n}\right)<-\frac{1}{2n}$$ And on adding such equations we get $$\log a_n-\log a_0<-\frac{1}{2}\sum_{i=1}^{n}\frac{1}{i}$$ Since the expression on RHS diverges to $-\infty$ it follows that $\log a_n\to-\infty $ and therefore $a_n\to 0$. Since $a_n$ is a subsequence of a convergent sequence $u_n$ it follows that $u_n\to 0$.
From the recurrence relation for $u_n$ it follows that it is decreasing and since $u_n\to 0$ by Leibniz test the alternating series $\sum (-1)^nu_n$ converges.
Next note that $$\sum_{i=0}^{n}(-1)^iu_i=\int_{0}^{\pi/2}\sum_{i=0}^{n}(-1)^{i}\sin^ix\,dx$$ The integral on right side can be written as  $$\int_{0}^{\pi/2}\frac{dx}{1+\sin x} +(-1)^{n+1}R_n$$ where $$0\leq R_n=\int_{0}^{\pi/2}\frac{\sin^{n+1}x}{1+\sin x} \, dx\leq \int_{0}^{\pi/2}\sin^{n+1}x\,dx=u_{n+1}$$ and therefore by squeeze theorem $R_n\to 0 $. And thus $(-1)^{n+1}R_n\to 0$. It follows that $$\sum_{n=0}^{\infty} (-1)^nu_n=\int_{0}^{\pi/2}\frac{dx}{1+\sin x} $$ The integral above can be easily evaluated by using standard substitution $t=\tan (x/2)$ and the integral value is $1$.
The final problem is the divergence of $\sum u_n$ which is easily handled by noting that $$u_n=\int_{0}^{1}\frac{t^n}{\sqrt{1-t^2}}\,dt\geq \int_{0}^{1}t^n\,dt=\frac{1}{n+1}$$ (also mentioned in comments to question by Kavi Rama Murthy). 
A: You can see the convergence of $u_n$ by induction:
$$\int^{\pi/2}_0 \cos x=[ \sin x]^{\pi/2}_0=1$$
$$\int^{\pi/2}_0 \cos^2 x=[ x/2+(1/4)\sin 2x]^{\pi/2}_0=\frac{\pi}{4}$$
$$\int^{\pi/2}_0 \cos^3 x=[ \sin x-(1/3)\sin^3 x]^{\pi/2}_0=1-\frac{1}{3}$$
$$\int^{\pi/2}_0 \cos^4 x=[(3/8)x+(1/32)\sin 4x+(1/16) \sin 2x]^{\pi/2}_0=\frac{3\pi}{16}$$
$$\int^{\pi/2}_0 \cos^5 x=[ \sin x -(1/5)\sin ^5x+..]^{\pi/2}_0=1-\frac{1}{5}. . .$$
$$\int^{\pi/2}_0 \cos^6 x=\frac{1}{192}[60x+48\sin 2x +9\sin 4x+4\sin^3 2x]^{\pi/2}_0=\frac{15\pi}{48}$$
Now if you sum these terms you can see the sum diverges.
Or we may use hint for 4; we have:
$$\int^{\pi/2}_0 \frac{dx}{1+\sin x}=\int^{\pi/2}_0(1-\sin x +\sin^2 x-\sin^3 x+ . . .= \Sigma\int^{\pi/2}_0(-1)^n \sin^n x=1 $$
Because:
$\sin x=\cos (\frac{\pi}{2}-x)=\cos 2(\frac{\pi}{4}-\frac{x}{2})$
$1+\sin x= 1+\cos 2(\frac{\pi}{4}-\frac{x}{2})= 2\cos^2 (\frac{\pi}{4}-\frac{x}{2})$
$$\int^{\pi/2}_0 \frac{dx}{1+\sin x}=\frac{1}{2}\int^{\pi/2}_0 \frac{dx}{\cos^2 (\frac{\pi}{4}-\frac{x}{2})}=[\tan (\frac{\pi}{4}-\frac{x}{2})]^{\pi/2}_0=1$$
That is the sum of $(-1)^n u_n$ converges to 1 and $u_n$ diverges.
$\int^{\pi/2}_0 \frac{dx}{1+\cos x}$
$\cos x= \cos 2(\frac{x}{2})$
$1+\cos x=1+\cos 2(\frac{x}{2})=2\cos^2 (\frac{x}{2})$
$$\int^{\pi/2}_0 \frac{dx}{1+\cos x}=\frac{1}{2}\int^{\pi/2}_0 \frac{dx}{\cos^2 x/2}=[\tan x/2]^{\pi/2}_0=1$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
u_{n} & \equiv \int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x =
\int_{0}^{\pi/2}\cos^{n}\pars{x}\,\dd x =
\int_{0}^{\pi/2}\exp\pars{n\ln\pars{\cos\pars{x}}}\,\dd x
\end{align}

With
  Laplace Method:

\begin{align}
&u_{n}  
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\int_{0}^{\infty}\expo{-nx^{2}/2}\pars{1 - {nx^{4} \over 12}}\,\dd x
\\[5mm] \implies &\
\bbx{u_{n} \equiv
\int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x \sim
\bbox[10px,#ffd]{\pars{{1 \over n^{1/2}} - {1 \over 4n^{3/2}}}\root{\pi \over 2}}
\quad\mbox{as}\quad n \to \infty}
\end{align}
