Mathematics

Find :
$$\int { \dfrac { { sec }^{ 2 }x }{ { tan }^{ 2 }x+4 } dx } $$


SOLUTION
$$\displaystyle\int{\dfrac{{\sec}^{2}{x}dx}{{\tan}^{2}{x}+4}}$$
Let $$t=\tan{x}\Rightarrow dt={\sec}^{2}{x}dx$$
$$=\displaystyle\int{\dfrac{dt}{{t}^{2}+4}}$$
$$=\dfrac{1}{2}{\tan}^{-1}{\dfrac{x}{2}}+c$$ using $$\displaystyle\int{\dfrac{dx}{{x}^{2}+{a}^{2}}}=\dfrac{1}{a}{\tan}^{-1}{\dfrac{x}{a}}+c$$
$$\therefore \displaystyle\int{\dfrac{{\sec}^{2}{x}dx}{{\tan}^{2}{x}+4}}=\dfrac{1}{2}{\tan}^{-1}{\dfrac{x}{2}}+c$$ where $$c$$ is the constant of integration.
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Subjective Medium Published on 17th 09, 2020
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