Question

The integral ∫ sec²/³ x cosec⁴/³ x dx is equal to: (Here C is a constant of integration)
(A) 3 tan⁻¹/³ x + C (B) -3/4 tan⁻⁴/³ x + C (C) -3 cot⁻¹/³ x + C (D) -3 tan⁻¹/³ x + C

Answer 1

∫ sec²/³ x cosec⁴/³ x dx is equal to

-3 tan⁻¹/³ x + C

option D is correct.

∫ sec⅔ x cosec⁴/³x dx

•changing secx and cosecx into cosx & sinx respectively

•we, get

∫ 1/(cos⅔x . sin⁴/³x)

•Divide both numerator and

denominator by cos²x

∫ (1/cos²x)/(cos^-⁴/³x . sin⁴/³x) dx

∫ (1/cos²x)/(cos^-⁴/³x . sin⁴/³x) dx

1/cos²x = sec²x &

(sin⁴/³x)/(cos⁴/³x) = tan⁴/³x

∫ (sec²x)/(tan⁴/³x) dx

•let tanx = t

sec²xdx = dt

∫ dt/(t⁴/³)

∫ t^-⁴/³dt

(t^-⅓)/(-1/3) + c

(-3)tan^-⅓x + c

hence option D is correct.

Answer 2

answer : option (D) -3 tan⁻¹/³ x + C

I = ∫sec⅔x. cosec⁴/³x dx

= ∫1/cos²x × (cos x. cosecx)⁴/³ dx

= ∫ sec²x. (cosx/sinx)⁴/³ dx

= ∫ sec²x/(tanx)⁴/³ dx

= ∫sec²x dx/(tanx)⁴/³ dx

let tanx = P

differentiating with respect to x

sec²x dx = dP

so, I = ∫ dP/(P)⁴/³

= ∫P-⁴/³ dP

= (P-¹/³)/(-4/3 + 1) + C

= -3(P)-¹/³ + C

putting P = tanx

so, I = -3tan-¹/³x + C

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