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
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