Question-156182

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} xdx}{12 \tan^{2} x - 4} = \int \frac{\sec^{2} x (1 + \tan^{2} x) dx}{12 \tan^{2} x - 4}

tanx = t

= \int \frac{(1 + t^{2}) dt}{{12 t}^{2} - 4} = \frac{1}{12} \int dt + \frac{4}{3} \int \frac{dt}{{12 t}^{2} - 4} = \frac{t}{12} + \frac{1}{6 \sqrt{3}} \ln ∣ \frac{2 \sqrt{3} t - 2}{2 \sqrt{3} t + 2} ∣

= \frac{\tan x}{12} + \frac{1}{6 \sqrt{3}} \ln ∣ \frac{\sqrt{3} \tan x - 1}{\sqrt{3} \tan x + 1} ∣ + C

Commented by MJS_new last updated on 10/Oct/21

Sir something went wrong

\frac{1 + t^{2}}{12 t^{2} - 4} = \frac{1}{12} + \frac{1}{9 t^{2} - 3}

\Rightarrow

\int \frac{1 + t^{2}}{12 t^{2} - 4} d t = \frac{t}{12} + \frac{\sqrt{3}}{18} \ln \frac{\sqrt{3} t - 1}{\sqrt{3} t + 1} \dots

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

Sorry sir . Thank you .

Commented by MJS_new last updated on 10/Oct/21

no reason to be sorry, {we}^{'} re here to help

each others

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