Question

Consider $I_1 = \int_0^{\frac{\pi}{2}}\frac{\cos x - \sin x}{1 - \sin x \cdot \cos x}dx$; $I_2 = \int_0^2 \log_e(\frac{2}{x} - 1)dx$; $I_3 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^5 x \ dx$; $I_4 = \int_{-\pi}^{\pi} \cos^3 x \ dx$.

Select the INCORRECT statements:

• A. The sum of all the integrals is zero
• B. The value of each integral is zero
• C. Value of exactly one integral is non-zero
• D. Value of at least one integral is non-zero

Answer 1

Answer: (C), (D)

Value of exactly one integral is non-zero, Value of at least one integral is non-zero

Answer 2

The problem requires us to evaluate four definite integrals and then determine which of the given statements are incorrect.

1. Evaluate $I_1 = \int_0^{\frac{\pi}{2}}\frac{\cos x - \sin x}{1 - \sin x \cdot \cos x}dx$

We use the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here $a=0, b=\frac{\pi}{2}$. So, $a+b-x = \frac{\pi}{2}-x$. $I_1 = \int_0^{\frac{\pi}{2}}\frac{\cos(\frac{\pi}{2}-x) - \sin(\frac{\pi}{2}-x)}{1 - \sin(\frac{\pi}{2}-x) \cdot \cos(\frac{\pi}{2}-x)}dx$ $I_1 = \int_0^{\frac{\pi}{2}}\frac{\sin x - \cos x}{1 - \cos x \cdot \sin x}dx$ $I_1 = -\int_0^{\frac{\pi}{2}}\frac{\cos x - \sin x}{1 - \sin x \cdot \cos x}dx$ This means $I_1 = -I_1$, which implies $2I_1 = 0$, so $I_1 = 0$.

2. Evaluate $I_2 = \int_0^2 \log_e(\frac{2}{x} - 1)dx$

We use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Here $a=2$. $I_2 = \int_0^2 \log_e(\frac{2}{2-x} - 1)dx$ $I_2 = \int_0^2 \log_e(\frac{2 - (2-x)}{2-x})dx$ $I_2 = \int_0^2 \log_e(\frac{x}{2-x})dx$ Now, add the original expression for $I_2$ and this new form: $2I_2 = \int_0^2 \left[ \log_e(\frac{2-x}{x}) + \log_e(\frac{x}{2-x}) \right] dx$ Using the logarithm property $\log A + \log B = \log(AB)$: $2I_2 = \int_0^2 \log_e\left( \frac{2-x}{x} \cdot \frac{x}{2-x} \right) dx$ $2I_2 = \int_0^2 \log_e(1) dx$ $2I_2 = \int_0^2 0 dx = 0$ So, $I_2 = 0$. (Note: This is an improper integral at $x=0$ and $x=2$, but it converges to 0.)

3. Evaluate $I_3 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^5 x \ dx$

We check if the integrand $f(x) = \sin^5 x$ is an even or odd function. $f(-x) = (\sin(-x))^5 = (-\sin x)^5 = -\sin^5 x = -f(x)$. Since $f(-x) = -f(x)$, $f(x)$ is an odd function. For an odd function, $\int_{-a}^a f(x) dx = 0$. Therefore, $I_3 = 0$.

4. Evaluate $I_4 = \int_{-\pi}^{\pi} \cos^3 x \ dx$

We check if the integrand $f(x) = \cos^3 x$ is an even or odd function. $f(-x) = (\cos(-x))^3 = (\cos x)^3 = \cos^3 x = f(x)$. Since $f(-x) = f(x)$, $f(x)$ is an even function. For an even function, $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$. So, $I_4 = 2\int_0^{\pi} \cos^3 x \ dx$. We can rewrite $\cos^3 x$ as $\cos^2 x \cdot \cos x = (1-\sin^2 x)\cos x$. Let $u = \sin x$. Then $du = \cos x dx$. When $x=0$, $u=\sin 0 = 0$. When $x=\pi$, $u=\sin \pi = 0$. So, $I_4 = 2\int_0^0 (1-u^2) du$. Since the upper and lower limits of integration are the same, the integral is 0. Therefore, $I_4 = 0$.

Summary of integral values: $I_1 = 0$ $I_2 = 0$ $I_3 = 0$ $I_4 = 0$

Evaluate the given statements:

• The sum of all the integrals is zero: $I_1 + I_2 + I_3 + I_4 = 0 + 0 + 0 + 0 = 0$. This statement is CORRECT.

• The value of each integral is zero: As calculated, $I_1=0, I_2=0, I_3=0, I_4=0$. This statement is CORRECT.

• Value of exactly one integral is non-zero: Since all integrals are zero, this statement is INCORRECT.

• Value of at least one integral is non-zero: Since all integrals are zero, this statement is INCORRECT.

The question asks to select the INCORRECT statements.

The final answer is $\boxed{\text{Value of exactly one integral is non-zero, Value of at least one integral is non-zero}}$

Explanation of the solution: Each integral is evaluated using properties of definite integrals.

1. $I_1$: Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$, the integral transforms into its negative, leading to $I_1=0$.
2. $I_2$: Using the same property as $I_1$, the integral transforms into $\int_0^2 \log_e(\frac{x}{2-x})dx$. Adding this to the original $I_2$ yields $\int_0^2 \log_e(1)dx = 0$, hence $I_2=0$.
3. $I_3$: The integrand $\sin^5 x$ is an odd function, and the integration limits are symmetric about zero ($[-\frac{\pi}{2}, \frac{\pi}{2}]$). For an odd function $f(x)$, $\int_{-a}^a f(x)dx = 0$. Thus, $I_3=0$.
4. $I_4$: The integrand $\cos^3 x$ is an even function, and the integration limits are symmetric about zero ($[-\pi, \pi]$). For an even function $f(x)$, $\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx$. Substituting $u=\sin x$ in $2\int_0^\pi (1-\sin^2 x)\cos x dx$ changes the limits of integration to $0$ to $0$, resulting in $I_4=0$. Since all integrals $I_1, I_2, I_3, I_4$ are zero, the statements "The sum of all the integrals is zero" and "The value of each integral is zero" are correct. The statements "Value of exactly one integral is non-zero" and "Value of at least one integral is non-zero" are incorrect.