Mathematics

# $I= \int \frac{1}{12+13cosx}dx.$

##### SOLUTION
$I = \int \frac{dx}{12+13cosx}$
Put $tan x/2 = t$
$\Rightarrow cosx = \frac{1-t^{2}}{1+t^{2}}$
$dt = \frac{2dt}{1+t^{2}}$
$\Rightarrow I = \int \frac{2dt}{(1+t^{2})(12+13\frac{(1-t^{2})}{1+t^{2}})} = \frac{2dt}{12(1+t^{2})+13(1-t^{2})}$
$= \int \frac{2dt}{25-t^{2}} = 2\times \frac{1}{2\times 5}$
$ln \left | \frac{5+t}{5-t} \right |+c$
$= \frac{1}{5}ln\left | \frac{5+tanx/2}{5-tanx/2} \right | +4$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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