Integral Calculus Question 264
Question: $ \int_{{}}^{{}}{\frac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}=} $
[Roorkee 1976; RPET 1991]
Options:
A) $ \tan x+\cot x+c $
B) $ \cot x-\tan x+c $
C) $ \tan x-\cot x+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}=\int_{{}}^{{}}{\frac{({{\cos }^{2}}x+{{\sin }^{2}}x),dx}{{{\cos }^{2}}x{{\sin }^{2}}x}} $ $ =\int_{{}}^{{}}{cose{c^{2}}x,dx}+\int_{{}}^{{}}{{{\sec }^{2}}x,dx}=-\cot x+\tan x+c $ .
BETA
