[tex] \large \bold{PROBLEM:}[/tex]
,[tex]\displaystyle \sf\Omega(n)=\sum_{k=2}^{n-1} \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2 k+1} x \cos ^{k} x+\sin ^{k} x \cos ^{2 k+1} x}{\sin ^{3 k+3} x+\cos ^{3 k+3} x} [/tex]Then...
fiind the omega
[tex] \Large \bold{SOLUTION:} [/tex]
[tex] \small \begin{array}{l} \displaystyle \textsf{Let }I = \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2 k+1} x \cos ^{k} x+\sin ^{k} x \cos ^{2 k+1} x}{\sin ^{3 k+3} x+\cos ^{3 k+3} x} dx \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle I = \int_{0}^{\frac{\pi}{4}} \frac{\sin^k x \cos^k x (\sin ^{k+1} x + \cos ^{k+1} x)}{(\sin ^{k+1} x + \cos ^{k+1} x)(\sin ^{2k+2} x - \sin ^{k+1} x \cos ^{k+1} x + \cos ^{2k+2} x)} dx \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle I = \int_{0}^{\frac{\pi}{4}} \frac{\sin^k x \cos^k x}{\sin ^{2k+2} x - \sin ^{k+1} x \cos ^{k+1} x + \cos ^{2k+2} x} dx \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle I = \int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin^k x \cos^k x}{\cos^{2k + 2}x}}{\frac{\sin ^{2k+2} x}{\cos^{2k + 2} x} - \frac{\sin ^{k+1} x \cos ^{k+1} x}{\cos^{2k + 2} x} + \frac{\cos ^{2k+2} x}{\cos^{2k + 2} x}} dx \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle I = \int_{0}^{\frac{\pi}{4}} \frac{\tan^k x \sec^2 x}{\tan^{2k+2} x - \tan^{k+1} x + 1} dx \end{array} [/tex]
[tex] \small \begin{array}{l} \begin{aligned} \quad \quad \textsf{Let }u &= \tan^{k+1} x \\ du &= (k + 1) \tan^k x \sec^2 x \: dx \\ \dfrac{du}{k + 1} &=\tan^k x \sec^2 x \: dx \end{aligned} \\ \\ \textsf{At }x= 0, u = \tan^{k + 1} (0) = 0 \\ \\ \textsf{At }x = \dfrac{\pi}{4}, u = \tan^{k + 1} \left(\dfrac{\pi}{4}\right) = 1^{k+1} = 1 \\ \\ \\ \displaystyle I = \frac{1}{k + 1} \int_{0}^{1} \frac{du}{u^2 - u + 1} \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle I= \frac{1}{k + 1}\int_{0}^{1} \frac{du}{u^2 - u + \frac{1}{4} + \frac{3}{4}} \\ \\ \displaystyle I = \frac{1}{k + 1} \int_{0}^{1} \frac{du}{\left(u - \frac{1}{2}\right)^2 + \big(\frac{\sqrt{3}}{2}\big)^2} \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle I = \Bigg[\frac{1}{k + 1}\bigg(\frac{\:2}{\sqrt{3}}\bigg)\arctan\left(\frac{u - \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\Bigg ]_0^1 \\ \\ \displaystyle I = \bigg[\frac{1}{k + 1}\bigg(\frac{\:2}{\sqrt{3}}\bigg)\arctan\left(\frac{2u - 1}{\sqrt{3}}\right)\bigg]_0^1 \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle I = \frac{1}{k + 1}\bigg(\frac{\:2}{\sqrt{3}}\bigg)\left(\arctan\bigg(\frac{1}{\sqrt{3}}\bigg) - \arctan\bigg(-\frac{1}{\sqrt{3}}\bigg)\right) \\ \\ \displaystyle I = \frac{1}{k+1}\left(\frac{2}{\sqrt{3}}\right)\!\left(\dfrac{\pi}{3}\right) = \frac{1}{k+1}\left(\frac{2\pi}{3\sqrt{3}}\right) \\ \: \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle \implies \Omega (n) = \frac{2\pi}{3\sqrt{3}} \sum_{k=2}^{n-1} \frac{1}{k+1} \\ \\ \displaystyle \implies \Omega (n) = \frac{2\pi}{3\sqrt{3}}\bigg(H_n - \bigg(1 + \frac{1}{2}\bigg)\bigg) \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle \implies \Omega (n) = \frac{2\pi}{3\sqrt{3}}H_n - \frac{\pi}{\sqrt{3}} \\ \\ \quad \textsf{where }H_n\textsf{ is the }nth\textsf{ harmonic number} \\ \\ \: \end{array} [/tex]
[tex] \small \begin{array}{l} \displaystyle \lim _{n \rightarrow \infty} \frac{\Omega (n)}{n} = \frac{2\pi}{3\sqrt{3}} \lim _{n \rightarrow \infty} \frac{H_n}{n} - \lim _{n \to \infty} \frac{\pi}{n\sqrt{3}} \\ \\ \: \end{array} [/tex]
[tex] \small \begin{array}{l} \underline{\bold{By\ Stolz-Cesàro\ Theorem,}} \\ \\ \displaystyle \quad \lim_{n\to \infty} \dfrac{H_{n+1} - H_n}{(n + 1) - n} = \lim_{n\to \infty} \dfrac{1 }{n + 1} = 0 \\ \\ \displaystyle \quad \implies \lim_{n\to \infty} \frac{H_n}{n} = 0 \\ \\ \\ \therefore \displaystyle \lim _{n \to \infty} \frac{\Omega (n)}{n} = 0 - 0 = \boxed{0}\quad\textit{Answer} \end{array} [/tex]
[tex] \large \bold{ANSWER:} \\ \huge \dashrightarrow{0}[/tex]
[tex] \large \color{indigo}{ \bold{Carry\: On\: Latex}}[/tex]
