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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals5 Marks
Question
Integrate the function: $\frac{1}{{1 - \tan x}}$
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Answer

Let $I = \int {\frac{1}{{1 - \tan x}}} dx$

$= \int {\frac{1}{{1 - \frac{{\sin x}}{{\cos x}}}}} dx$

$= \int {\frac{1}{{\left( {\frac{{\cos x - \sin x}}{{\cos x}}} \right)}}} dx$

$= \int {\frac{{\cos x}}{{\cos x - \sin x}}} dx$

$= \frac{1}{2}\int {\frac{{2\cos x}}{{\cos x - \sin x}}} dx$

$= \frac{1}{2}\int {\frac{{\cos x + \cos x}}{{\cos x - \sin x}}} dx$

Adding and subtracting sin x in the numerator,

$= \frac{1}{2}\int {\frac{{\cos x - \sin x + \sin x + \cos x}}{{\cos x - \sin x}}} dx$

$= \frac{1}{2}\int {\frac{{\left( {\cos x - \sin x} \right) + \left( {\sin x + \cos x} \right)}}{{\cos x - \sin x}}} dx$

$= \frac{1}{2}\int {\frac{{\cos x - \sin x}}{{\cos x - \sin x}}} + \frac{{\sin x + \cos x}}{{\cos x - \sin x}}dx$

$= \frac{1}{2}\int {\left( {1 + \frac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right)} dx$

$= \frac{1}{2}\left[ {\int {1dx-\int {\frac{{ - \sin x - \cos x}}{{\cos x - \sin x}}} } } \right]dx$

$= \frac{1}{2}\left[ {x - \log \left| {\cos x - \sin x} \right|} \right] + c$ $[\because \int \frac{f'(x)}{f(x)} dx=log|f(x)|+c]$

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