logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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

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

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $(i)\;\int sin xdx=-\cos x$.
  • $(ii)\;\int\sec^2xdx=\tan x.$
$\int\frac{2-3\sin x}{\cos^2x}dx.$
 
$\;\;\;=\int(\frac{2}{\cos^2x}-\frac{3\sin x}{\cos^2x})dx.[But\frac{1}{\cos^2x}=\sec^2x$ 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
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...