Integral Calculus Question 429
Question: $ \int_{{}}^{{}}{\frac{cose{c^{2}}x}{1+\cot x}dx=} $
[MNR 1973]
Options:
A) $ \log (1+\cot x)+c $
B) $ -\log (1+\cot x)+c $
C) $ \frac{1}{2{{(1+\cot x)}^{2}}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ 1+\cot x=t\Rightarrow cose{c^{2}}x,dx=-dt, $ then $ \int_{{}}^{{}}{\frac{cose{c^{2}}x}{1+\cot x},dx}=-\int_{{}}^{{}}{\frac{1}{t},dt=-\log t+c}=-\log (1+\cot x)+c $ .
BETA
