Evaluate ∫ 0 2 π sin 2 n ( x ) d x ∫ 0 2 π sin 2 n + 2 ( x ) d x for all positive integer n .
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A quicker way would be to put n=2 and verify the options
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1 . −
n ≥ 1 , ∫ 0 2 π sin 2 n + 1 ( x ) d x = ∫ 0 2 π sin 2 n ( x ) ⋅ sin ( x ) d x = Integrating by parts, = ( − cos ( x ) ⋅ sin 2 n ( x ) ) 0 π / 2 + 2 n ⋅ ∫ 0 2 π sin 2 n − 1 ( x ) ⋅ cos 2 ( x ) d x = ( cos 2 ( x ) = 1 − sin 2 ( x ) ) 2 n ⋅ ( ∫ 0 2 π sin 2 n − 1 ( x ) d x − ∫ 0 2 π sin 2 n + 1 ( x ) d x ) ⇒ Let's call I = ∫ 0 2 π sin 2 n + 1 ( x ) d x I = ∫ 0 2 π sin 2 n + 1 ( x ) d x = 2 n + 1 2 n ⋅ ∫ 0 2 π sin 2 n − 1 ( x ) d x ⇒ ∫ 0 2 π sin 2 n ( x ) d x ∫ 0 2 π sin 2 n + 2 ( x ) d x = 2 n + 2 2 n + 1 = 2 n + 2 2 n + 2 − 1 = 1 − 2 n + 2 1
2 . −
You can also use Beta function as suggested by @Aditya Sharma
∫ 0 2 π sin 2 n + 2 ( x ) d x = ∫ 0 2 π sin 2 ( n + 3 / 2 ) − 1 ( x ) ⋅ cos 2 ( 1 / 2 ) − 1 ( x ) d x = 2 B ( n + 3 / 2 , 1 / 2 ) = 2 ⋅ Γ ( n + 2 ) Γ ( n + 3 / 2 ) ⋅ Γ ( 1 / 2 )
In the same way, ∫ 0 2 π sin 2 n ( x ) d x = 2 B ( n + 1 / 2 , 1 / 2 ) = 2 ⋅ Γ ( n + 1 ) Γ ( n + 1 / 2 ) ⋅ Γ ( 1 / 2 )
This implies that, ∫ 0 2 π sin 2 n ( x ) d x ∫ 0 2 π sin 2 n + 2 ( x ) d x = 2 ⋅ Γ ( n + 1 ) Γ ( n + 1 / 2 ) ⋅ Γ ( 1 / 2 ) 2 ⋅ Γ ( n + 2 ) Γ ( n + 3 / 2 ) ⋅ Γ ( 1 / 2 ) = ( n + 1 ) ( n + 1 / 2 ) = ( n + 1 ) ( n + 1 − 1 / 2 ) = 1 − 2 n + 2 1