Question

If $f(x) = \int_{0}^{\frac{\pi}{2}} \frac{\ln(1+x\sin^2\theta)}{\sin^2\theta} d\theta, x\ge 0$, then $\frac{\pi}{f'(3)}$ equals

Answer 1

4

Answer 2

Let the given function be $f(x) = \int_{0}^{\frac{\pi}{2}} \frac{\ln(1+x\sin^2\theta)}{\sin^2\theta} d\theta$ for $x \ge 0$. We need to find the value of $\frac{\pi}{f'(3)}$.

First, we find the derivative of $f(x)$ with respect to $x$ using Leibniz rule for differentiation under the integral sign. The rule states that if $f(x) = \int_{a}^{b} h(x, \theta) d\theta$, where $a$ and $b$ are constants, then $f'(x) = \int_{a}^{b} \frac{\partial}{\partial x} h(x, \theta) d\theta$. In our case, $h(x, \theta) = \frac{\ln(1+x\sin^2\theta)}{\sin^2\theta}$, $a=0$, and $b=\frac{\pi}{2}$. The partial derivative of $h(x, \theta)$ with respect to $x$ is: $\frac{\partial}{\partial x} \left(\frac{\ln(1+x\sin^2\theta)}{\sin^2\theta}\right) = \frac{1}{\sin^2\theta} \cdot \frac{\partial}{\partial x} (\ln(1+x\sin^2\theta))$ Using the chain rule, $\frac{\partial}{\partial x} (\ln(1+x\sin^2\theta)) = \frac{1}{1+x\sin^2\theta} \cdot \frac{\partial}{\partial x} (1+x\sin^2\theta) = \frac{1}{1+x\sin^2\theta} \cdot \sin^2\theta$. So, $\frac{\partial}{\partial x} h(x, \theta) = \frac{1}{\sin^2\theta} \cdot \frac{\sin^2\theta}{1+x\sin^2\theta} = \frac{1}{1+x\sin^2\theta}$. Thus, $f'(x) = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+x\sin^2\theta} d\theta$.

Now, we need to evaluate this integral. Let $I(x) = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+x\sin^2\theta} d\theta$. We can divide the numerator and denominator by $\cos^2\theta$: $I(x) = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2\theta}{\sec^2\theta + x\tan^2\theta} d\theta$. Using the identity $\sec^2\theta = 1+\tan^2\theta$: $I(x) = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2\theta}{1+\tan^2\theta + x\tan^2\theta} d\theta = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2\theta}{1+(1+x)\tan^2\theta} d\theta$. Let $u = \tan\theta$. Then $du = \sec^2\theta d\theta$. When $\theta = 0$, $u = \tan(0) = 0$. When $\theta = \frac{\pi}{2}$, $u = \tan(\frac{\pi}{2}) = \infty$. The integral becomes: $I(x) = \int_{0}^{\infty} \frac{du}{1+(1+x)u^2}$. We can write the denominator as $1+(\sqrt{1+x}u)^2$. Let $v = \sqrt{1+x}u$. Then $dv = \sqrt{1+x}du$, so $du = \frac{dv}{\sqrt{1+x}}$. When $u = 0$, $v = 0$. When $u = \infty$, $v = \infty$. The integral becomes: $I(x) = \int_{0}^{\infty} \frac{1}{1+v^2} \frac{dv}{\sqrt{1+x}} = \frac{1}{\sqrt{1+x}} \int_{0}^{\infty} \frac{dv}{1+v^2}$. The integral $\int_{0}^{\infty} \frac{dv}{1+v^2}$ is a standard integral: $\int_{0}^{\infty} \frac{dv}{1+v^2} = [\arctan(v)]_{0}^{\infty} = \arctan(\infty) - \arctan(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. So, $I(x) = \frac{1}{\sqrt{1+x}} \cdot \frac{\pi}{2} = \frac{\pi}{2\sqrt{1+x}}$. Thus, $f'(x) = \frac{\pi}{2\sqrt{1+x}}$.

We need to find the value of $\frac{\pi}{f'(3)}$. First, we find $f'(3)$ by substituting $x=3$ into the expression for $f'(x)$: $f'(3) = \frac{\pi}{2\sqrt{1+3}} = \frac{\pi}{2\sqrt{4}} = \frac{\pi}{2 \cdot 2} = \frac{\pi}{4}$.

Finally, we calculate $\frac{\pi}{f'(3)}$: $\frac{\pi}{f'(3)} = \frac{\pi}{\frac{\pi}{4}} = \pi \cdot \frac{4}{\pi} = 4$.

The final answer is $\boxed{4}$.

Explanation:

1. Use Leibniz rule to differentiate $f(x)$ under the integral sign.
2. The derivative $f'(x)$ is found to be $\int_{0}^{\frac{\pi}{2}} \frac{1}{1+x\sin^2\theta} d\theta$.
3. Evaluate the integral by transforming it using trigonometric identities and substitution ($\tan\theta$).
4. The integral evaluates to $\frac{\pi}{2\sqrt{1+x}}$, so $f'(x) = \frac{\pi}{2\sqrt{1+x}}$.
5. Calculate $f'(3)$ by substituting $x=3$ into the expression for $f'(x)$.
6. $f'(3) = \frac{\pi}{4}$.
7. Calculate the required value $\frac{\pi}{f'(3)} = \frac{\pi}{\frac{\pi}{4}} = 4$.