Please answer this problem: the integral involving trigonometric functions?

in Calculus Answers by Level 1 User (120 points)

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1 Answer

tan(x/3)+cot(x/3)=

sin(x/3)/cos(x/3)+cos(x/3)/sin(x/3)=

(sin²(x/3)+cos²(x/3))/(sin(x/3)cos(x/3))=

2/sin(2x/3).

So ∫4csc²(2x/3)dx is the same integral=

-4cot(2x/3)×(3/2)=-6cot(2x/3)=-6/tan(2x/3)=

-6(1-tan²(x/3))/(2tan(x/3)))=

3(tan²(x/3)-1)/tan(x/3)=3tan(x/3)-3cot(x/3).

Now add in the constant of integration:

3tan(x/3)-3cot(x/3)+c, answer option III.

by Top Rated User (788k points)

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