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Question Number 170926 by Tawa11 last updated on 04/Jun/22

$$\int \frac{\mathrm{dx}}{{\sin }^2\mathrm{x}-6\sin \left(\mathrm{x}\right)+5}$$

Answered by thfchristopher last updated on 04/Jun/22

$$\int \frac{dx}{{\sin }^{2}x-6\sin x+5}$$$$\mathrm{Let}t=\tan \frac{x}{2}\Rightarrow \sin x=\frac{2t}{1+t^2}$$$$dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx\Rightarrow dx=\frac{2dt}{1+t^2}$$$$\therefore \int \frac{dx}{{\sin }^{2}x-6\sin x+5}$$$$=\int \frac{\frac{2dt}{1+t^2}}{{\left(\frac{2t}{1+t^2}\right)}^2-6\left(\frac{2t}{1+t^2}\right)+5}$$$$=\int \frac{2(1+t^2)}{4t^2-12t(1+t^2)+5{(1+t^2)}^2}dt$$$$=\int \frac{2(1+t^2)}{5t^4-12t^3+14t^2-12t+5}dt$$$$=\int \frac{2(1+t^2)}{{(t-1)}^2(5t^2-2t+5)}dt$$$$=\frac{1}{2}\int [\frac{1}{{(t-1)}^2}-\frac{1}{5t^2-2t+5}]dt$$$$=\frac{1}{2}\int \frac{dt}{{(t-1)}^2}-\frac{1}{10}\int \frac{dt}{t^2-\frac{2}{5}t+1}$$$$=\frac{1}{2(1-t)}-\frac{1}{10}\int \frac{dt}{{(t-\frac{1}{5})}^2+{\left(\frac{2\sqrt{6}}{5}\right)}^2}$$$$\mathrm{For}\int \frac{dt}{{(t-\frac{1}{5})}^2+{\left(\frac{2\sqrt{6}}{5}\right)}^2},$$$$\mathrm{Let}t-\frac{1}{5}=\frac{2\sqrt{6}}{5}\tan u,$$$$dt=\frac{2\sqrt{6}}{5}{\sec }^{2}udu$$$$\therefore \int \frac{dt}{{(t-\frac{1}{5})}^2+{\left(\frac{2\sqrt{6}}{5}\right)}^2}$$$$=\frac{5}{2\sqrt{6}}\int \frac{\cancel{{\sec }^{2}u}}{\cancel{{\sec }^{2}u}}du$$$$=\frac{5}{2\sqrt{6}}\int du$$$$=\frac{5}{2\sqrt{6}}u+C$$$$=\frac{5}{2\sqrt{6}}{\tan }^{-1}\left(\frac{5t-1}{2\sqrt{6}}\right)+C$$$$\mathrm{Hence},$$$$\int \frac{dx}{{\sin }^{2}x-6\sin x+5}$$$$=\frac{1}{2(1-t)}-\frac{1}{4\sqrt{6}}{\tan }^{-1}\left(\frac{5t-1}{2\sqrt{6}}\right)+C$$$$=\frac{1}{2[1-\tan \frac{x}{2}]}-\frac{1}{4\sqrt{6}}{\tan }^{-1}\left[\frac{5\tan \left(\frac{x}{2}\right)-1}{2\sqrt{6}}\right]+C$$

Commented by Tawa11 last updated on 04/Jun/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by peter frank last updated on 05/Jun/22

$$\mathrm{thank}\mathrm{you}$$

Answered by mr W last updated on 05/Jun/22

$$=\int \frac{dx}{(\sin x-1)(\sin x-5)}$$$$=\frac{1}{4}(\int \frac{dx}{\sin x-5}-\int \frac{dx}{\sin x-1})$$$$=\frac{1}{4}(-\frac{1}{\sqrt{6}}{\tan }^{-1}\frac{5\tan \frac{x}{2}-1}{2\sqrt{6}}-\frac{2}{\tan \frac{x}{2}-1})+C$$

Commented by Tawa11 last updated on 05/Jun/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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