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
Correct Answer: B
Solution :
Put \[1+\cot x=t\Rightarrow \text{cose}{{\text{c}}^{2}}x\,dx=-dt,\] then \[\int_{{}}^{{}}{\frac{\text{cose}{{\text{c}}^{2}}x}{1+\cot x}\,dx}=-\int_{{}}^{{}}{\frac{1}{t}\,dt=-\log t+c}=-\log (1+\cot x)+c\].You need to login to perform this action.
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