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Question Number 156305 by cortano last updated on 10/Oct/21

$$\int \frac{\sec {}^2\left(\mathrm{x}\right)}{3\sin {}^2\left(\mathrm{x}\right)-1}\mathrm{dx}$$

Answered by puissant last updated on 10/Oct/21

$$\mathrm{\Omega }=\int \frac{{sec}^{2}x}{3{sin}^{2}x-1}dx=\int \frac{{sec}^{4}x}{3{tan}^{2}x-{sec}^{2}x}dx$$$$u=tanx\to du={sec}^{2}xdx=(1+u^2)du$$$$\Rightarrow \mathrm{\Omega }=\int \frac{1+u^2}{3u^2-(1+u^2)}du=\int \frac{1+u^2}{2u^2-1}du$$$$\mathrm{\Omega }=\int \frac{1}{2u^2-1}du+\int \frac{u^2}{2u^2-1}du$$$$=\frac{1}{2}\int \frac{du}{u^2-\frac{1}{2}}+\frac{1}{2}\int \frac{2u^2-1+1}{2u^2-1}du$$$$=\frac{1}{2}\int \frac{du}{u^2-\frac{1}{2}}+\frac{1}{4}\int \frac{du}{u^2-\frac{1}{2}}+\frac{1}{2}\int du=\frac{3}{4}\int \frac{du}{u^2-\frac{1}{2}}+\frac{1}{2}\int du$$$$=-\frac{6}{4}\int \frac{du}{1-\left(\frac{u}{\sqrt{2}}\right)}-\frac{u}{2}=-\frac{6\sqrt{2}}{4}argtanh\left(\sqrt{2}u\right)-\frac{u}{2}+C..$$$$$$$$\therefore \because \mathrm{\Omega }=-\frac{3\sqrt{2}}{2}argtanh\left(\sqrt{2}tanx\right)-\frac{tanx}{2}+C$$

Commented by cortano last updated on 10/Oct/21

$$\mathrm{ok}.\mathrm{i}\mathrm{got}\mathrm{in}\tan \mathrm{x}\mathrm{form}.$$

Commented by puissant last updated on 10/Oct/21

$$Okey...$$

Commented by puissant last updated on 10/Oct/21

$$MrCortano,anotherimportant$$$$formula:$$$$argth\left(x\right)+C=\frac{1}{2}ln\left(\frac{x+1}{x-1}\right)+C=\int \frac{1}{x^2-1}dx..$$

Commented by cortano last updated on 10/Oct/21

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

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