If f ( x ) = 3 1 + x 2
And g ( t ) = ∫ 0 t f ( x ) d x
Then find the value of ∫ ln ( tan ( 6 π ) ) ln ( tan ( 3 π ) ) g ( x ) d x .
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yeah bro solved in the same way!!! nice problem BTW.
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f ( x ) = 3 1 + x 2
f ( − x ) = 3 1 + x 2
f ( x ) = f ( − x ) ⋯ ⋯ ⋯ ( t h e r e f o r e f ( x ) i s e v e n f u n c t i o n )
g ( t ) = ∫ 0 t f ( x ) d x ⋯ ⋯ ⋯ ⋯ ( t h e r e f o r e g ( t ) i s o d d f u n c t i o n )
l e t ∫ l n ( t a n ( 6 π ) ) l n ( t a n ( 3 π ) ) g ( x ) d x = I
∴ I = ∫ − l n ( 3 ) l n ( 3 ) g ( x ) d x
∵ ∫ − a a f ( x ) d x = 0 ⋯ ⋯ ⋯ ( w h e r e f ( x ) i s a o d d f u n t i o n )
∴ I = 0