Integrating Squares

Calculus Level 2

Knowing that sin 2 x + cos 2 x = 1 \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 = π 2 \displaystyle \int_0^{\pi /2} \sin^2 x \, dx + \int_0^{\pi /2} \cos^2 x \, dx = \dfrac\pi2 .
(B) : 0 π / 2 sin 2 x d x = π 4 \displaystyle \int_0^{\pi /2} \sin^2 x \, dx = \dfrac\pi4 .
(C) : 0 π / 2 cos 2 x d x = π 4 \displaystyle \int_0^{\pi /2} \cos^2 x \, dx = \dfrac\pi4 .

(A) only (B) only (C) only (A), (B) and (C)

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2 solutions

Pranshu Gaba
Sep 1, 2016

Since sin 2 x \sin^2 x and cos 2 x \cos ^2 x are both bounded in 0 x π 2 0 \le x \le \frac\pi2 , 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 \int_{0} ^{\pi/2} \sin^2 x \, dx + \int _{0} ^{\pi/2} \cos^2 x \, dx = \int _{0} ^{\pi/2} (\sin^2 x + \cos^2 x ) \, dx Since sin 2 x + cos 2 x \sin ^2 x + \cos ^2 x is equal to 1 1 for all x x , the integral simplifies to 0 π / 2 1 d x = π 2 \displaystyle \int _{0} ^{\pi/2} 1\, dx = \frac\pi 2 .

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 \int_0 ^ a f(x) \, dx = \int_0 ^ a f(a - x) \, dx

0 π / 2 sin 2 x d x = 0 π / 2 sin 2 ( π 2 x ) d x = 0 π / 2 cos 2 x d x \begin{aligned} \int_{0} ^{\pi/2} \sin^2 x \, dx & = \int_{0} ^{\pi/2} \sin^2 (\frac{\pi}{2} - x) \, dx \\ & = \int_{0} ^{\pi/2} \cos ^2 x \, dx \end{aligned}

Since 0 π / 2 sin 2 x d x = 0 π / 2 cos 2 x d x \int_{0} ^{\pi/2} \sin^2 x \, dx = \int_{0} ^{\pi/2} \cos^2 x \, dx , and their sum is equal to π 2 \frac{\pi}{2} , each of them is equal to π 4 \frac\pi4


Alternatively, we could use the double angle identity cos 2 x = 1 2 sin 2 x = 2 cos 2 x 1 \cos 2x = 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 1 cos 2 x 2 d x = x sin 2 x 2 2 0 π / 2 = π 4 \int _{0} ^{\pi/2} \sin^2 x \, dx = \int _{0} ^{\pi/2} \frac{1 - \cos 2x}{2} \, dx = \left. \frac{x - \frac{\sin 2x}{2}}{2} \right| _{0} ^{\pi/2} = \frac\pi4

0 π / 2 cos 2 x d x = 0 π / 2 1 + cos 2 x 2 d x = x + sin 2 x 2 2 0 π / 2 = π 4 \int _{0} ^{\pi/2} \cos^2 x \, dx = \int _{0} ^{\pi/2} \frac{1 + \cos 2x}{2} \, dx = \left. \frac{x +\frac{\sin 2x}{2}}{2} \right| _{0} ^{\pi/2} = \frac\pi4

A = 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 = 0 π 2 d x = π 2 \begin{aligned} A & = \int_0^\frac \pi 2 \sin^2 x \ dx + \int_0^\frac \pi 2 \cos^2 x \ dx \\ & = \int_0^\frac \pi 2 \sin^2 x + \cos^2 x \ dx \\ & = \int_0^\frac \pi 2 \ dx = \frac \pi 2 \end{aligned}

Using the identity a b f ( x ) d x = a b f ( a + b x ) d x \displaystyle \int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx , we have:

B = 0 π 2 sin 2 x d x = 1 2 0 π 2 sin 2 x + sin 2 ( π 2 x ) d x = 1 2 0 π 2 sin 2 x + cos 2 x d x = π 4 \begin{aligned} B & = \int_0^\frac \pi 2 \sin^2 x \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \sin^2 x + \sin^2 \left(\frac \pi 2 - x\right) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \sin^2 x + \cos^2 x \ dx = \frac \pi 4 \end{aligned}

Similarly, C = 0 π 2 cos 2 x d x = π 4 C = \displaystyle \int_0^\frac \pi 2 \cos^2 x \ dx = \frac \pi 4 .

Therefore, (A), (B) and (C) \boxed{\text{(A), (B) and (C)}} are true.

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