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Question Number 175305 by cortano1 last updated on 26/Aug/22

Answered by mr W last updated on 27/Aug/22

$$lett=\frac{x}{6}$$$$6{\int }_0^{\frac{\pi }{36}}\frac{1-\cos 2t}{\sin 3t}dt$$$$=12{\int }_0^{\frac{\pi }{36}}\frac{{\sin }^{2}t}{\sin t(3-4{\sin }^{2}t)}dt$$$$=12{\int }_0^{\frac{\pi }{36}}\frac{\sin t}{4{\cos }^{2}t-1}dt$$$$=12{\int }_{\frac{\pi }{36}}^0\frac{d\left(\cos t\right)}{(2\cos t+1)(2\cos t-1)}$$$$=12{\int }_{\cos \frac{\pi }{36}}^1\frac{du}{(2u+1)(2u-1)}$$$$=6{\int }_{\cos \frac{\pi }{36}}^1(\frac{du}{2u-1}-\frac{du}{2u+1})$$$$=3{\left[\ln \frac{2u-1}{2u+1}\right]}_{\cos \frac{\pi }{36}}^1$$$$=3[\ln \frac{1}{3}-\ln \frac{2\cos \frac{\pi }{36}-1}{2\cos \frac{\pi }{36}+1}]$$$$=3\ln \frac{2\cos \frac{\pi }{36}+1}{3(2\cos \frac{\pi }{36}-1)}$$

Commented by cortano1 last updated on 27/Aug/22

$$yes===thanksyou$$

Commented by Tawa11 last updated on 27/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

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