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Question Number 216425 by universe last updated on 07/Feb/25

Commented by universe last updated on 07/Feb/25

$$findsum$$

Answered by MrGaster last updated on 08/Feb/25

$$={\int }_0^{\frac{\pi }{2}}\underset{n=1}{\sum ^{10}}\frac{{\sin }^2\left(3^{n}x\right)}{\cos \left(3^{n+1}x\right)}dx$$$$={\int }_0^{\frac{\pi }{2}}(\frac{{\sin }^2\left(3x\right)}{\cos \left(9x\right)}+\frac{{\sin }^2\left(9x\right)}{\cos \left(27x\right)}+\dots +\frac{{\sin }^2\left(3^{10}x\right)}{\cos \left(3^{11}x\right)})dx$$$$=\underset{n=1}{\sum ^{10}}{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2\left(3^{n}x\right)}{\cos \left({}^{3n+1}x\right)}dx$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}{\int }_0^{3^n\frac{\pi }{2}}\frac{{\sin }^{2}u}{\cos \left(3u\right)}du\right]$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}{\int }_0^{3^n\frac{\pi }{2}}\frac{{\sin }^{2}u}{\cos \left(3u\right)}du\right]$$$$=\underset{n=1}{\sum ^{10}}[\frac{1}{3^n}{(\frac{1}{6}\ln \mid \frac{1+\sin \left(3u\right)}{1-\sin (3u}\mid -\frac{1}{3}\tan \left(3u\right)]}_0^{3^n\frac{\pi }{2}}$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}(\frac{1}{6}\ln \mid \frac{1+\sin \left(3^{n+1}\frac{\pi }{2}\right)}{1-\sin \left(3^{n+1}\frac{\pi }{2}\right)}\mid -\frac{1}{3}\tan \left(3^{n+1}\frac{\pi }{2}\right))\right]$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}(\frac{1}{6}\ln \mid \frac{1+{(-1)}^{n+1}}{1-{(-1)}^{n+1}}\mid -\frac{1}{3}\tan \left(3^{n+1}\frac{\pi }{2}\right))\right]$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}(\frac{1}{6}\ln \mid \frac{2}{‘‘0^{″}}\mid -\frac{1}{3}\tan \left(3^{n+1}\frac{\pi }{2}\right))\right]$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}(\frac{1}{6}\ln \mid ‘‘{\mathrm{\infty }}^{″}\mid -\frac{1}{3}\tan \left(3^{n+1}\frac{\pi }{2}\right))\right]$$$$\mathrm{Next}\mathrm{some}‘‘{\mathrm{illegal}}^{″}\mathrm{operations}\mathrm{will}$$$$\mathrm{becarried}\mathrm{out}:$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}(\frac{\mathrm{\infty }}{6}-\frac{\mathrm{\infty }}{3})\right]$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}(\frac{\mathrm{\infty }}{6}-\frac{2\mathrm{\infty }}{6})\right]$$$$=\underset{n=1}{\sum ^{10}}\left[\frac{1}{3^n}(-\frac{\mathrm{\infty }}{6})\right]$$$$=\underset{n=1}{\sum ^{10}}[-\frac{\mathrm{\infty }}{6\cdot 3^n}]$$$$=-\frac{\mathrm{\infty }}{6}\underset{n=1}{\sum ^{10}}\frac{1}{3^n}$$$$=\frac{\mathrm{\infty }}{6}\cdot \frac{1}{2}(1-\frac{1}{3^{10}})$$$$=-\frac{\mathrm{\infty }}{12}(1-\frac{1}{3^{10}})$$$$\overline{)\begin{array}{c}-\frac{\mathrm{\infty }}{12}\stackrel{?}{\approx }‘‘0.9113919361034865649028795807632075543645{\dots }^{″}\end{array}}$$

Answered by MathematicalUser2357 last updated on 25/Feb/25

$$\downarrow \downarrow \downarrow$$

Commented by MathematicalUser2357 last updated on 25/Feb/25

sin^2(3 x) sec(9 x) + sin^2(9 x) sec(27 x) + sin^2(27 x) sec(81 x) + sin^2(81 x) sec(243 x) + sin^2(243 x) sec(729 x) + sin^2(729 x) sec(2187 x) + sin^2(2187 x) sec(6561 x) + sin^2(6561 x) sec(19683 x) + sin^2(19683 x) sec(59049 x) + sin^2(59049 x) sec(177147 x)

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