The integral ∫(from π/6 to π/3) tan³x · sin²3x(2sec²x · sin²3x + 3tanx · sin6x)dx is
The integral ∫(from π/6 to π/3) tan³x · sin²3x(2sec²x · sin²3x + 3tanx · sin6x)dx is
Option 1 -
-1/18
Option 2 -
-1/9
Option 3 -
1/9
Option 4 -
1/18
-
1 Answer
-
Correct Option - 1
Detailed Solution:∫ [π/6 to π/3] (d/dx (tan? x) * sin³3x + tan? x * d/dx (sin³3x) dx
= 1/2 ∫ [π/6 to π/3] d/dx (tan? x * sin?3x) dx
= 1/2 [tan? x · sin?3x] from π/6 to π/3
= 1/2 [ (√3)? · 0 - 1/ (√3)? )]
= -1/18
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