Question

Find:

$I = \displaystyle \sf \int \dfrac{x( tan\ x + 2 \ tan \ 2x + 4 \ tan \ 4x)}{cot \ x - 8 \ cot\ 8x} \ dx$

Answer 1

Solution:

Before we solve this problem, we must get to know some trigonometric identities as follows

tan2x = 2 tanx / (1 - tan²x)

• tanx - cotx

= 1/cotx - cotx

= (1 - cot²x) / cotx

= - 2 (cot²x - 1) / (2 cotx)

= - 2 cot2x

• cot2x - tan2x

= cot2x - 1/cot2x

= 2 (cot²2x - 1) / (2 cot2x)

= 2 cot4x

• cot4x - tan4x

= cot4x - 1/cot4x

= 2 (cot²4x - 1) / (2 cot4x)

= 2 cot8x

• tanx + 2 tan2x + 4 tan4x - cotx + 8 cot8x

= (tanx - cotx) + 2 tan2x + 4 tan4x + 8 cot8x

= - 2 cot2x + 2 tan2x + 4 tan4x + 8 cot8x

= - 2 (cot2x - tan2x) + 4 tan4x + 8 cot8x

= - 2 (2 cot4x) + 4 tan4x + 8 cot8x

= - 4 cot4x + 4 tan4x + 8 cot8x

= - 4 (cot4x - tan4x) + 8 cot8x

= - 4 (2 cot8x) + 8 cot8x

= - 8 cot8x + 8 cot8x

= 0

or, tanx + 2 tan2x + 4 tan4x - cotx + 8 cot8x = 0

or, tanx + 2 tan2x + 4 tan4x = cotx - 8 cot8x

•••

∴ I = ∫ x (tanx + 2 tan2x + 4 tan4x) dx / (cotx - 8 cot8x)

= ∫ x (cotx - 8 cot8x) dx / (cotx - 8 cot8x)

= ∫ x dx

= x²/2 + c

where c = constant of integration

Therefore, the required integral is

x²/2 + c

Answer 2

Answer:

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