Question

If $I = \int_{0}^{2\pi} \sin^2x dx$, then

• A. $I = 2\int_{0}^{\pi} \sin^2x dx$
• B. $I = 4\int_{0}^{\pi/2} \sin^2x dx$
• C. $I = \int_{0}^{2\pi} \cos^2x dx$
• D. $I = 8\int_{0}^{\pi/4} \sin^2x dx$

Answer 1

Answer: (A), (B), (C)

I = 2\int_{0}^{\pi} sin^2x dx, I = 4\int_{0}^{\pi/2} sin^2x dx, I = \int_{0}^{2\pi} cos^2x dx

Answer 2

The problem asks us to identify the correct statements regarding the integral $I = \int_{0}^{2\pi} \sin^2x dx$.

First, let's evaluate the integral $I$: We use the trigonometric identity $\sin^2x = \frac{1 - \cos(2x)}{2}$. $I = \int_{0}^{2\pi} \frac{1 - \cos(2x)}{2} dx$ $I = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{2\pi}$ $I = \frac{1}{2} \left[ \left( 2\pi - \frac{\sin(4\pi)}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \right]$ Since $\sin(4\pi) = 0$ and $\sin(0) = 0$: $I = \frac{1}{2} \left[ (2\pi - 0) - (0 - 0) \right] = \frac{1}{2} (2\pi) = \pi$ So, $I = \pi$.

Now, let's check each given option:

Option 1: $I = 2\int_{0}^{\pi} \sin^2x dx$ We use the property of definite integrals: $\int_{0}^{2a} f(x) dx = 2\int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$. Here, $f(x) = \sin^2x$ and $2a = 2\pi$, so $a = \pi$. Let's check $f(2\pi-x)$: $f(2\pi-x) = \sin^2(2\pi-x) = (-\sin x)^2 = \sin^2x = f(x)$. Since $f(2\pi-x) = f(x)$, the property applies. Therefore, $I = \int_{0}^{2\pi} \sin^2x dx = 2\int_{0}^{\pi} \sin^2x dx$. This option is correct.

Option 2: $I = 4\int_{0}^{\pi/2} \sin^2x dx$ From Option 1, we have $I = 2\int_{0}^{\pi} \sin^2x dx$. Now, consider the integral $\int_{0}^{\pi} \sin^2x dx$. We apply the same property again: $\int_{0}^{2a} f(x) dx = 2\int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$. Here, $f(x) = \sin^2x$ and $2a = \pi$, so $a = \pi/2$. Let's check $f(\pi-x)$: $f(\pi-x) = \sin^2(\pi-x) = (\sin x)^2 = \sin^2x = f(x)$. Since $f(\pi-x) = f(x)$, the property applies. Thus, $\int_{0}^{\pi} \sin^2x dx = 2\int_{0}^{\pi/2} \sin^2x dx$. Substitute this back into the expression for $I$: $I = 2 \left( 2\int_{0}^{\pi/2} \sin^2x dx \right) = 4\int_{0}^{\pi/2} \sin^2x dx$. This option is correct.

Option 3: $I = \int_{0}^{2\pi} \cos^2x dx$ Let's evaluate $\int_{0}^{2\pi} \cos^2x dx$. We use the trigonometric identity $\cos^2x = \frac{1 + \cos(2x)}{2}$. $\int_{0}^{2\pi} \cos^2x dx = \int_{0}^{2\pi} \frac{1 + \cos(2x)}{2} dx$ $= \frac{1}{2} \left[ x + \frac{\sin(2x)}{2} \right]_{0}^{2\pi}$ $= \frac{1}{2} \left[ \left( 2\pi + \frac{\sin(4\pi)}{2} \right) - \left( 0 + \frac{\sin(0)}{2} \right) \right]$ $= \frac{1}{2} \left[ (2\pi + 0) - (0 + 0) \right] = \frac{1}{2} (2\pi) = \pi$ Since $I = \pi$ and $\int_{0}^{2\pi} \cos^2x dx = \pi$, this option is correct. Alternatively, we know $\sin^2x + \cos^2x = 1$. $\int_{0}^{2\pi} (\sin^2x + \cos^2x) dx = \int_{0}^{2\pi} 1 dx = [x]_{0}^{2\pi} = 2\pi$. So, $\int_{0}^{2\pi} \sin^2x dx + \int_{0}^{2\pi} \cos^2x dx = 2\pi$. $I + \int_{0}^{2\pi} \cos^2x dx = 2\pi$. Since $I = \pi$, we have $\pi + \int_{0}^{2\pi} \cos^2x dx = 2\pi$, which implies $\int_{0}^{2\pi} \cos^2x dx = \pi$. Thus, $I = \int_{0}^{2\pi} \cos^2x dx$.

Option 4: $I = 8\int_{0}^{\pi/4} \sin^2x dx$ Let's evaluate $8\int_{0}^{\pi/4} \sin^2x dx$: $8\int_{0}^{\pi/4} \frac{1 - \cos(2x)}{2} dx = 4\int_{0}^{\pi/4} (1 - \cos(2x)) dx$ $= 4 \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{\pi/4}$ $= 4 \left[ \left( \frac{\pi}{4} - \frac{\sin(\pi/2)}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \right]$ $= 4 \left[ \left( \frac{\pi}{4} - \frac{1}{2} \right) - (0 - 0) \right]$ $= 4 \left( \frac{\pi}{4} - \frac{1}{2} \right) = \pi - 2$ Since $I = \pi$ and $\pi - 2 \neq \pi$, this option is incorrect.

All options 1, 2, and 3 are correct.