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Question Number 156182 by cortano last updated on 09/Oct/21

Answered by PRITHWISH SEN 2 last updated on 10/Oct/21

$$\int \frac{{\sec }^4\mathrm{xdx}}{12\tan {}^2\mathrm{x}-4}=\int \frac{{\sec }^2\mathrm{x}(1+\tan {}^2\mathrm{x})\mathrm{dx}}{12\tan {}^2\mathrm{x}-4}$$$$\mathrm{tanx}=\mathrm{t}$$$$=\int \frac{(1+{\mathrm{t}}^2)\mathrm{dt}}{{12\mathrm{t}}^2-4}=\frac{1}{12}\int \mathrm{dt}+\frac{4}{3}\int \frac{\mathrm{dt}}{{12\mathrm{t}}^2-4}=\frac{\mathrm{t}}{12}+\frac{1}{6\sqrt{3}}\ln \mid \frac{2\sqrt{3}\mathrm{t}-2}{2\sqrt{3}\mathrm{t}+2}\mid$$$$=\frac{\tan \mathrm{x}}{12}+\frac{1}{6\sqrt{3}}\ln \mid \frac{\sqrt{3}\tan \mathrm{x}-1}{\sqrt{3}\tan \mathrm{x}+1}\mid +\mathrm{C}$$$$$$$$$$$$$$

Commented by MJS_new last updated on 10/Oct/21

$$\mathrm{Sir}\mathrm{something}\mathrm{went}\mathrm{wrong}$$$$\frac{1+t^2}{12t^2-4}=\frac{1}{12}+\frac{1}{9t^2-3}$$$$\Rightarrow$$$$\int \frac{1+t^2}{12t^2-4}dt=\frac{t}{12}+\frac{\sqrt{3}}{18}\ln \frac{\sqrt{3}t-1}{\sqrt{3}t+1}...$$

Commented by PRITHWISH SEN 2 last updated on 10/Oct/21

$$\mathrm{Sorry}\mathrm{sir}.\mathrm{Thank}\mathrm{you}.$$

Commented by MJS_new last updated on 10/Oct/21

$$\mathrm{no}\mathrm{reason}\mathrm{to}\mathrm{be}\mathrm{sorry},{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{here}\mathrm{to}\mathrm{help}$$$$\mathrm{each}\mathrm{others}$$

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