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Question:

The integral ∫sec²/³x cosec⁴/³x dx is equal to?

-3/4tan⁻⁴/³x+C

-3tan⁻¹/³x+C

3tan⁻¹/³x+C

-3cot⁻¹/³x+C

Solution:

Correct option is D. -3tan⁻¹/³x+CI=∫dx(sin x)⁴/³(cos x)²/³I=∫dx(sin x cos x)⁴/³cos²x ⇒I=∫sec²x(tan x)⁴/³dxPut tan x=t ⇒sec²xdx=dt∴I=∫dt t⁴/³ ⇒I=-3t⁻¹/³+c ⇒I=-3(tan x)⁻¹/³+c.