The problem consists of four parts: a) Express $25\cosh x - 24\sinh x$ in the form $R\cosh(x-\alpha)$, where $R > 0$ and $\alpha > 0$. b) If $f(x) = 25\cosh x - 24\sinh x$, find the critical number of $f$ and classify it. c) Let $a, b \ge 0$. If $f$ is continuous on $\mathbb{R}$ and $\int_a^b f(t) dt = 10$, find $\int_{\sqrt{b}}^{\sqrt{a}} xf(x^2) dx$. d) Evaluate the integral $\int \cos \theta \sin^2 \theta d\theta$ using the substitution $u = \sin \theta$.

1. Problem Description

The problem consists of four parts:

a) Express $25\cosh x - 24\sinh x$ in the form $R\cosh(x-\alpha)$, where $R > 0$ and $\alpha > 0$.

b) If $f(x) = 25\cosh x - 24\sinh x$, find the critical number of $f$ and classify it.

c) Let $a, b \ge 0$. If $f$ is continuous on $\mathbb{R}$ and $\int_a^b f(t) dt = 10$, find $\int_{\sqrt{b}}^{\sqrt{a}} xf(x^2) dx$.

d) Evaluate the integral $\int \cos \theta \sin^2 \theta d\theta$ using the substitution $u = \sin \theta$.

2. Solution Steps

a)

We are given the identity $\cosh(A \pm B) = \cosh A \cosh B \pm \sinh A \sinh B$.

We want to express $25\cosh x - 24\sinh x$ in the form $R\cosh(x-\alpha)$.

$R\cosh(x-\alpha) = R(\cosh x \cosh \alpha - \sinh x \sinh \alpha) = R\cosh \alpha \cosh x - R\sinh \alpha \sinh x$.

Equating the coefficients of $\cosh x$ and $\sinh x$, we get:

$R\cosh \alpha = 25$

$R\sinh \alpha = 24$

Dividing the second equation by the first, we get:

$\tanh \alpha = \frac{R\sinh \alpha}{R\cosh \alpha} = \frac{24}{25}$

$\alpha = \tanh^{-1}\left(\frac{24}{25}\right)$

Squaring both equations and subtracting the second squared equation from the first squared equation, we get:

$R^2\cosh^2 \alpha - R^2\sinh^2 \alpha = 25^2 - 24^2$

$R^2(\cosh^2 \alpha - \sinh^2 \alpha) = 625 - 576$

$R^2(1) = 49$

$R = \sqrt{49} = 7$ (since $R > 0$)

So, $R = 7$ and $\alpha = \tanh^{-1}\left(\frac{24}{25}\right)$.

b)

$f(x) = 25\cosh x - 24\sinh x = 7\cosh(x-\alpha)$, where $\alpha = \tanh^{-1}\left(\frac{24}{25}\right)$.

To find the critical number, we need to find the derivative of $f(x)$ and set it to

0. $f'(x) = 7\sinh(x-\alpha)$

Setting $f'(x) = 0$, we get:

$7\sinh(x-\alpha) = 0$

$\sinh(x-\alpha) = 0$

$x - \alpha = 0$

$x = \alpha = \tanh^{-1}\left(\frac{24}{25}\right)$

To classify the critical number, we find the second derivative:

$f''(x) = 7\cosh(x-\alpha)$

$f''(\alpha) = 7\cosh(\alpha - \alpha) = 7\cosh(0) = 7(1) = 7 > 0$

Since $f''(\alpha) > 0$, the critical number is a local minimum.

c)

Let $u = x^2$. Then $du = 2x dx$, so $x dx = \frac{1}{2}du$.

When $x = \sqrt{b}$, $u = (\sqrt{b})^2 = b$.

When $x = \sqrt{a}$, $u = (\sqrt{a})^2 = a$.

So, the integral becomes:

$\int_{\sqrt{b}}^{\sqrt{a}} xf(x^2) dx = \int_b^a f(u) \frac{1}{2} du = \frac{1}{2}\int_b^a f(u) du = -\frac{1}{2}\int_a^b f(u) du$

Given that $\int_a^b f(t) dt = 10$, we have:

$-\frac{1}{2}\int_a^b f(u) du = -\frac{1}{2}(10) = -5$

d)

Given $\int \cos \theta \sin^2 \theta d\theta$ and $u = \sin \theta$, then $du = \cos \theta d\theta$.

So, the integral becomes:

$\int \cos \theta \sin^2 \theta d\theta = \int u^2 du = \frac{1}{3}u^3 + C = \frac{1}{3}\sin^3 \theta + C$

3. Final Answer

a) $R=7$, $\alpha = \tanh^{-1}\left(\frac{24}{25}\right)$

b) Critical number: $x = \tanh^{-1}\left(\frac{24}{25}\right)$, Local minimum

c) $-5$

d) $\frac{1}{3}\sin^3 \theta + C$