Question
Download Solution PDFमूल्यांकन करें: \(\smallint \frac{{1 - \cos 2{\rm{x}}}}{{1 - {{\sin }^2}{\rm{x}}}}{\rm{\;dx}}\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFधारणा:
1 - cos 2x = 2 sin2 x
1 – sin2 x = cos2 x
\(\smallint {\sec ^2}{\rm{xdx}} = \tan {\rm{x}} + {\rm{c}}\)
गणना:
माना कि I = \(\smallint \frac{{1 - \cos 2{\rm{x}}}}{{1 - {{\sin }^2}{\rm{x}}}}{\rm{\;dx}}\)
\( = \smallint \frac{{2{{\sin }^2}{\rm{x}}}}{{{{\cos }^2}{\rm{x}}}}{\rm{\;dx}}\)
\( = 2\smallint {\tan ^2}{\rm{xdx}}\)
\( = 2\smallint \left( {{{\sec }^2}{\rm{x}} - 1} \right){\rm{dx}}\)
= 2 [tan x – x] + c
= 2 tan x – 2x + c
Last updated on Jun 11, 2025
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