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Question Number 134641 by benjo_mathlover last updated on 06/Mar/21

$$\mathcal{B}=\int \frac{1-\sin 6x}{1+\sin 6x}dx$$

Answered by Dwaipayan Shikari last updated on 06/Mar/21

$$\int \frac{2}{1+sin6x}-1dx6x=u$$$$=\frac{1}{3}\int \frac{1}{1+sinu}du-x$$$$=\frac{2}{3}\int \frac{1}{1+\frac{2t}{1+t^2}}.\frac{1}{1+t^2}dt-xtan\frac{u}{2}=t$$$$=\frac{2}{3}\int \frac{1}{{(1+t)}^2}dt-x=-\frac{2}{3(1+t)}-x+C=-\frac{2}{3(1+tan3x)}-x+C$$

Answered by Olaf last updated on 06/Mar/21

$$\mathrm{By}\mathrm{using}\sin 6x=\frac{2\tan \left(3x\right)}{1+{\tan }^2\left(3x\right)}$$$$\mathrm{and}\mathrm{integrating}\mathrm{by}\mathrm{parts}:$$$$\mathcal{B}=-\frac{2}{3}.\frac{1}{1+\tan \left(3x\right)}-x+\mathrm{C}$$

Answered by EDWIN88 last updated on 06/Mar/21

$$\mathcal{B}=\int \frac{1-\sin 6\mathrm{x}}{1+\sin 6\mathrm{x}}\mathrm{dx}=\int \tan {}^2(\frac{\pi }{4}-3\mathrm{x})\mathrm{dx}$$$$\mathcal{B}=\int [\sec {}^2(\frac{\pi }{4}-3\mathrm{x})-1]\mathrm{dx}$$$$\mathcal{B}=-\frac{1}{3}\tan (\frac{\pi }{4}-3\mathrm{x})-\mathrm{x}+\mathrm{c}$$

Answered by Ar Brandon last updated on 06/Mar/21

$$\mathcal{B}=\int \frac{1-\sin 6\mathrm{x}}{1+\sin 6\mathrm{x}}\mathrm{dx}=\int \frac{{(1-\sin 6\mathrm{x})}^2}{{\cos }^{2}6\mathrm{x}}\mathrm{dx}$$$$=\int \frac{1-2\sin 6\mathrm{x}+{\sin }^{2}6\mathrm{x}}{{\cos }^{2}6\mathrm{x}}\mathrm{dx}=\int \frac{2-2\sin 6\mathrm{x}-{\cos }^{2}6\mathrm{x}}{{\cos }^{2}6\mathrm{x}}$$$$=\frac{\tan 6\mathrm{x}}{3}-\frac{1}{3\cos 6\mathrm{x}}-\mathrm{x}+\mathcal{C}$$

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