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प्रश्न
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
विकल्प
tan x − x + C
x + tan x + C
x − tan x + C
− x − cot x + C
MCQ
उत्तर
x − tan x + C
\[\int\left( \frac{\cos 2x - 1}{\cos 2x + 1} \right)dx\]
\[ = \int\left( \frac{1 - 2 \sin^2 x - 1}{2 \cos^2 x - 1 + 1} \right)dx\]
\[ = - \int \tan^2 x dx\]
\[ = - \int\left( \sec^2 x - 1 \right)dx\]
\[ = \int\left( 1 - \sec^2 x \right)dx\]
\[ = x - \tan x + C\]
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