Question

Let $y=f(x)$ be a differentiable function satisfying $\int_2^x f(t) dt + 2 = \frac{x^2}{2} + \int_x^2 t^2 f(t) dt$ then \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(x)+x^9-x^3+x+1}{\cos^2 x} dx = \text{_______}.

Answer 1

2

Answer 2

1. Find $f(x)$: The given equation is $\int_2^x f(t) dt + 2 = \frac{x^2}{2} + \int_x^2 t^2 f(t) dt$. Rewrite the equation as $\int_2^x f(t) dt + \int_2^x t^2 f(t) dt = \frac{x^2}{2} - 2$. Differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus: $f(x) + x^2 f(x) = x$. $(1+x^2)f(x) = x$. Thus, $f(x) = \frac{x}{1+x^2}$.

2. Evaluate the definite integral: The integral to evaluate is $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(x)+x^9-x^3+x+1}{\cos^2 x} dx$. Substitute $f(x) = \frac{x}{1+x^2}$: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{x}{1+x^2}+x^9-x^3+x+1}{\cos^2 x} dx$. Let the numerator be $N(x) = \frac{x}{1+x^2}+x^9-x^3+x+1$ and the denominator be $D(x) = \cos^2 x$. The interval of integration is $[-\frac{\pi}{4}, \frac{\pi}{4}]$, which is symmetric about $x=0$.

3. Analyze the parity of the integrand: The denominator $D(x) = \cos^2 x$ is an even function since $\cos^2(-x) = \cos^2 x$. Let's analyze the numerator $N(x)$. We know $f(x) = \frac{x}{1+x^2}$ is an odd function since $f(-x) = \frac{-x}{1+(-x)^2} = -f(x)$. Let's check $N(-x)$: $N(-x) = f(-x) + (-x)^9 - (-x)^3 + (-x) + 1$ $N(-x) = -f(x) - x^9 + x^3 - x + 1$. Now, consider $N(x) + N(-x)$: $N(x) + N(-x) = \left(\frac{x}{1+x^2}+x^9-x^3+x+1\right) + \left(-\frac{x}{1+x^2}-x^9+x^3-x+1\right)$ $N(x) + N(-x) = 2$. This means $N(x) = \frac{N(x)+N(-x)}{2} + \frac{N(x)-N(-x)}{2} = \frac{2}{2} + \frac{N(x)-N(-x)}{2} = 1 + \frac{N(x)-N(-x)}{2}$. The term $\frac{N(x)-N(-x)}{2}$ is an odd function: $\frac{N(x)-N(-x)}{2} = \frac{1}{2} \left[\left(\frac{x}{1+x^2}+x^9-x^3+x+1\right) - \left(-\frac{x}{1+x^2}-x^9+x^3-x+1\right)\right]$ $= \frac{1}{2} \left[\frac{2x}{1+x^2} + 2x^9 - 2x^3 + 2x\right] = \frac{x}{1+x^2} + x^9 - x^3 + x$. This is indeed an odd function. So, $N(x) = 1 + (\text{an odd function})$.

   The integrand is $\frac{N(x)}{D(x)} = \frac{1 + (\text{odd function})}{\cos^2 x} = \frac{1}{\cos^2 x} + \frac{\text{odd function}}{\cos^2 x}$. The term $\frac{\text{odd function}}{\cos^2 x}$ is an odd function (odd/even = odd). The integral of an odd function over a symmetric interval $[-a, a]$ is 0. So, $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\text{odd function}}{\cos^2 x} dx = 0$.

4. Calculate the remaining integral: We are left with $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\cos^2 x} dx$. $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x dx$. The antiderivative of $\sec^2 x$ is $\tan x$. $I = [\tan x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \tan\left(\frac{\pi}{4}\right) - \tan\left(-\frac{\pi}{4}\right)$. Since $\tan\left(\frac{\pi}{4}\right) = 1$ and $\tan\left(-\frac{\pi}{4}\right) = -1$, $I = 1 - (-1) = 1 + 1 = 2$.

The value of the integral is 2.