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Find the integral $\int\frac{\large 2-3\sin x}{\large \cos^2x}dx$

$\begin{array}{1 1}2\tan x-3\sec x+c \\ 2\cot x-3\sec x+c \\ 3\tan x-2\sec x+c \\ 3\cot x-2\sec x+c \end{array}$

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  • $(i)\;\int sin xdx=-\cos x$.
  • $(ii)\;\int\sec^2xdx=\tan x.$
$\int\frac{2-3\sin x}{\cos^2x}dx.$
$\;\;\;=\int(\frac{2}{\cos^2x}-\frac{3\sin x}{\cos^2x})dx.[But\frac{1}{\cos^2x}=\sec^2x$ and $\frac{\sin x}{\cos^2x}=\sec xtan x.$]
$\;\;\;=\int2\sec^2xdx-3\sec x\tan xdx$
$\;\;\;=2\int\sec^2xdx-3\int \sec xtan xdx.$
$\;\;\;=2\tan x-3\sec x+c.$
Hence $\int\frac{2-3\sin x}{\cos^2x}dx=2\tan x-3\sec x+c.$


answered Jan 27, 2013 by sreemathi.v
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