Question Detail

MTH 102 Practice Questions MTH 102 Objective Question

Question

∫cot(√3x) dx
Options
A -csc²√3x
B √3 ln|sin√3x|
C -sec√3x
D (1/√3)ln|sin√3x|
Correct Answer
Correct Answer

Option D is the correct answer.

Detailed Explanation

∫cot(u)du=ln|sinu|+C. With u=√3x, du=√3 dx, so ∫cot(√3x)dx=(1/√3)ln|sin√3x|+C.

Hint

Recall ∫cot(u)du = ln|sin u|, then adjust for the chain-rule factor √3.

Question Info