The question
For any integer n n n ∫ − π π e cos 2 x Sin 3 ( 2 n + 1 ) x d x \displaystyle\int_{-\pi}^{\pi} e^{\cos^2x}\operatorname{Sin}^3(2n + 1)x\,dx ∫ − π π e c o s 2 x Sin 3 ( 2 n + 1 ) x d x
(a) − 1 -1 − 1
(b) 0 0 0
(c) 1 1 1
(d) 2 2 2
Exam answer Show all explanations
Let f ( x ) = e cos 2 x sin 3 ( 2 n + 1 ) x f(x) = e^{\cos^2x}\sin^3(2n + 1)x f ( x ) = e c o s 2 x sin 3 ( 2 n + 1 ) x
Since cos ( − x ) = cos x \cos(-x) = \cos x cos ( − x ) = cos x sin ( − θ ) = − sin θ \sin(-\theta) = -\sin\theta sin ( − θ ) = − sin θ
f ( − x ) = e cos 2 x sin 3 ( − ( 2 n + 1 ) x ) = − e cos 2 x sin 3 ( 2 n + 1 ) x = − f ( x ) \begin{aligned} f(-x) &= e^{\cos^2x}\sin^3(-(2n + 1)x) \\ &= -e^{\cos^2x}\sin^3(2n + 1)x \\ &= -f(x) \end{aligned} f ( − x ) = e c o s 2 x sin 3 ( − ( 2 n + 1 ) x ) = − e c o s 2 x sin 3 ( 2 n + 1 ) x = − f ( x ) Why?
Replace every x x x − x -x − x
cos 2 ( − x ) = ( cos x ) 2 \cos^2(-x) = (\cos x)^2 cos 2 ( − x ) = ( cos x ) 2 e cos 2 x e^{\cos^2x} e c o s 2 x
The sine factor changes sign:
sin ( ( 2 n + 1 ) ( − x ) ) = − sin ( 2 n + 1 ) x \begin{aligned} &\sin((2n + 1)(-x)) \\ &= -\sin(2n + 1)x \end{aligned} sin (( 2 n + 1 ) ( − x )) = − sin ( 2 n + 1 ) x Cubing keeps the minus sign: ( − s ) 3 = − s 3 (-s)^3 = -s^3 ( − s ) 3 = − s 3
So f f f ∫ − a a f ( x ) d x = 0 \int_{-a}^{a}f(x)\,dx = 0 ∫ − a a f ( x ) d x = 0
Why?
For a continuous function with f ( − x ) = − f ( x ) f(-x) = -f(x) f ( − x ) = − f ( x ) − a -a − a a a a
∫ − a a f ( x ) d x = 0 \int_{-a}^{a}f(x)\,dx = 0 ∫ − a a f ( x ) d x = 0 Here a = π a = \pi a = π
∫ − π π e cos 2 x sin 3 ( 2 n + 1 ) x d x = 0 \int_{-\pi}^{\pi}e^{\cos^2x}\sin^3(2n + 1)x\,dx = 0 ∫ − π π e c o s 2 x sin 3 ( 2 n + 1 ) x d x = 0 Why?
Watch out: this holds for every integer n n n 2 n + 1 2n + 1 2 n + 1
Answer: (b) 0 0 0
Where this comes from Chapter 7, Integrals : Definite integrals: odd function property.
Question type: Limits from − a -a − a a a a
How to spot it: click to see Type 11. Limits from − a -a − a a a a
Spot it: the two limits are the same number with opposite signs.
Here: The limits are − π -\pi − π π \pi π − x -x − x e cos 2 x e^{\cos^2x} e c o s 2 x sin 3 ( 2 n + 1 ) x \sin^3(2n + 1)x sin 3 ( 2 n + 1 ) x
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