Mathematics

Evaluate: $$\displaystyle\int {{{2 - 3\sin x} \over {{{\cos }^2}x}}dx} $$


SOLUTION
Solution:
$$I=\displaystyle \int { \frac { 2-3\sin { x }  }{ \cos ^{ 2 }{ x }  } dx } $$

$$=\displaystyle \int { 2\sec ^{ 2 }{ x } dx } -3\int { \frac { \sin { x }  }{ \cos ^{ 2 }{ x }  } dx } $$

$$\displaystyle =2\tan { x } -3\int { \frac { \sin { x }  }{ \cos ^{ 2 }{ x }  } dx } $$

Let $$\cos x=t\quad \Rightarrow \ -\sin x dx=dt$$

$$\displaystyle I=2\tan { x } +3\int { \frac { 1 }{ { t }^{ 2 } } dx } $$

$$=2\tan x-\dfrac {3}{t}+c\ \Rightarrow \ 2\tan x-\dfrac {3}{\cos x}+c=2\tan x-3\sec x+c$$
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Subjective Medium Published on 17th 09, 2020
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