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evaluate the following integral ∫ tan 2x tan 3x tan 5x dx

evaluate the following integral
∫ tan 2x tan 3x tan 5x dx

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int tan(2x)tan(3x)tan(5x)dx
tan5x = tan(3x+2x) = \frac{ tan3x+tan2x}{1-tan(3x)tan(2x)}
tan3x+tan2x = tan5x-tan(2x)tan(3x)tan(5x)
Put in the integral
I = \int [tan5x - tan3x - tan2x]dx
I = \int tan5xdx - \int tan3xdx - \int tan2xdx
I = \frac{log(sec5x)}{5} - \int tan3xdx - \int tan2xdx
I = \frac{log(sec5x)}{5} - \frac{log(sec3x)}{3} - \int tan2xdx
I = \frac{log(sec5x)}{5} - \frac{log(sec3x)}{3} - \frac{log(sec2x)}{2}+c

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