Integral Calculus Question 308
Question: $ \int_{{}}^{{}}{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx=} $
[MP PET 1996]
Options:
A) $ \tan x+\cot x+c $
B) $ \tan x+cosec,x+c $
C) $ -\tan x+\cot x+c $
D) $ \tan x+\sec x+c $
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Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}},dx=\int_{{}}^{{}}{{{\sec }^{2}}x,dx-\int_{{}}^{{}}{cose{c^{2}},x,dx}} $ $ =\tan x+\cot x+c. $
BETA
