Knowing that sin 2 x + cos 2 x = 1 , then which of the following is true?
(A)
:
∫
0
π
/
2
sin
2
x
d
x
+
∫
0
π
/
2
cos
2
x
d
x
=
2
π
.
(B)
:
∫
0
π
/
2
sin
2
x
d
x
=
4
π
.
(C)
:
∫
0
π
/
2
cos
2
x
d
x
=
4
π
.
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A = ∫ 0 2 π sin 2 x d x + ∫ 0 2 π cos 2 x d x = ∫ 0 2 π sin 2 x + cos 2 x d x = ∫ 0 2 π d x = 2 π
Using the identity ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x , we have:
B = ∫ 0 2 π sin 2 x d x = 2 1 ∫ 0 2 π sin 2 x + sin 2 ( 2 π − x ) d x = 2 1 ∫ 0 2 π sin 2 x + cos 2 x d x = 4 π
Similarly, C = ∫ 0 2 π cos 2 x d x = 4 π .
Therefore, (A), (B) and (C) are true.
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Since sin 2 x and cos 2 x are both bounded in 0 ≤ x ≤ 2 π , we can say that
∫ 0 π / 2 sin 2 x d x + ∫ 0 π / 2 cos 2 x d x = ∫ 0 π / 2 ( sin 2 x + cos 2 x ) d x Since sin 2 x + cos 2 x is equal to 1 for all x , the integral simplifies to ∫ 0 π / 2 1 d x = 2 π .
To find the other two integrals, we will use the property of definite integrals ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x
∫ 0 π / 2 sin 2 x d x = ∫ 0 π / 2 sin 2 ( 2 π − x ) d x = ∫ 0 π / 2 cos 2 x d x
Since ∫ 0 π / 2 sin 2 x d x = ∫ 0 π / 2 cos 2 x d x , and their sum is equal to 2 π , each of them is equal to 4 π
Alternatively, we could use the double angle identity cos 2 x = 1 − 2 sin 2 x = 2 cos 2 x − 1 to evaluate the integrals.
∫ 0 π / 2 sin 2 x d x = ∫ 0 π / 2 2 1 − cos 2 x d x = 2 x − 2 sin 2 x ∣ ∣ ∣ ∣ ∣ 0 π / 2 = 4 π
∫ 0 π / 2 cos 2 x d x = ∫ 0 π / 2 2 1 + cos 2 x d x = 2 x + 2 sin 2 x ∣ ∣ ∣ ∣ ∣ 0 π / 2 = 4 π