Integral Calculus Question 110
Question: $ \int_{{}}^{{}}{\frac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}dx}= $
[MP PET 1993; BIT Ranchi 1982]
Options:
A) $ -\cot x-2x+c $
B) $ -2\cot x-2x+c $
C) $ -2\cot x-x+c $
D) $ -2\cot x+x+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}},dx=\int_{{}}^{{}}{(cose{c^{2}}x+{{\cot }^{2}}x),dx} $ $ =\int_{{}}^{{}}{(2cose{c^{2}}x-1),dx=-2\cot x-x+c.} $
BETA
