<s>[INST] 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}^*$.
  
  
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*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 $$u_n \geq \frac{1}{n + 1}$$




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*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.
 [/INST] 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.
$\int^{\pi/2}_0 \frac{dx}{1+\cos x}$
$\cos x= \cos 2(\frac{x}{2})$
$1+\cos x=1+\cos 2(\frac{x}{2})=2\cos^2 (\frac{x}{2})$
$$\int^{\pi/2}_0 \frac{dx}{1+\cos x}=\frac{1}{2}\int^{\pi/2}_0 \frac{dx}{\cos^2 x/2}=[\tan x/2]^{\pi/2}_0=1$$
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