Step 1: Simplifying the Integrand using Trigonometric Substitution

The integrand contains the inverse trigonometric expression $\sin^{-1} \left( \frac{2x}{1+x^2} \right)$. We utilize the substitution $x = \tan \theta$. Consequently, $dx = \sec^2 \theta \, d\theta$. [By the substitution rule for integrals].

Under this transformation, the expression simplifies using the double-angle identity for sine: $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$. Thus, $\sin^{-1} \left( \frac{2\tan\theta}{1+\tan^2\theta} \right) = \sin^{-1}(\sin 2\theta) = 2\theta$, provided $2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Given $x \in [0, 1]$, $\theta \in [0, \frac{\pi}{4}]$, satisfying the domain constraint.

Step 2: Adjusting Limits of Integration

We must transform the limits from $x$-space to $\theta$-space:

• When $x = 0$, $\tan \theta = 0 \implies \theta = 0$.
• When $x = 1$, $\tan \theta = 1 \implies \theta = \frac{\pi}{4}$.

Step 3: Setting up the Transformed Integral

Substituting the variables into the original integral $I = \int_{0}^{1} \sin^{-1} \left( \frac{2x}{1+x^2} \right) dx$:

$I = \int_{0}^{\pi/4} (2\theta) \sec^2 \theta \, d\theta = 2 \int_{0}^{\pi/4} \theta \sec^2 \theta \, d\theta$

Step 4: Integration by Parts

Applying the formula $\int u \, dv = uv - \int v \, du$, let $u = \theta$ and $dv = \sec^2 \theta \, d\theta$. Then $du = d\theta$ and $v = \tan \theta$. [Per Integration by Parts theorem].

$I = 2 \left[ \theta \tan \theta - \int \tan \theta \, d\theta \right]_{0}^{\pi/4}$

$I = 2 \left[ \theta \tan \theta - \ln|\sec \theta| \right]_{0}^{\pi/4}$

Step 5: Evaluating at the Limits

Applying the Fundamental Theorem of Calculus:

$I = 2 \left[ \left( \frac{\pi}{4} \tan \frac{\pi}{4} - \ln|\sec \frac{\pi}{4}| \right) - (0 \cdot \tan 0 - \ln|\sec 0|) \right]$

Since $\tan \frac{\pi}{4} = 1$, $\sec \frac{\pi}{4} = \sqrt{2}$, $\tan 0 = 0$, and $\sec 0 = 1$ (noting $\ln(1) = 0$):

$I = 2 \left[ \frac{\pi}{4}(1) - \ln(\sqrt{2}) - 0 \right]$

$I = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \ln(2) \right]$

$I = \frac{\pi}{2} - \ln(2)$

Final Solution: $\frac{\pi}{2} - \ln(2)$

Step 1: Substitution and Variable Transformation

Given the integral: $I = \int_{0}^{2} x \sqrt{x+2} dx$

As suggested, let $x + 2 = t^2$. [By the Method of Substitution for Definite Integrals]

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(x+2) = \frac{d}{dx}(t^2) \implies dx = 2t dt$

We must also transform the variable $x$ in terms of $t$:

$x = t^2 - 2$

Step 2: Determining New Limits of Integration

When $x$ is the lower limit $x = 0$:

$0 + 2 = t^2 \implies t^2 = 2 \implies t = \sqrt{2}$ (taking the positive root for the principal branch)

When $x$ is the upper limit $x = 2$:

$2 + 2 = t^2 \implies t^2 = 4 \implies t = 2$

Step 3: Constructing the Transformed Integral

Substituting $x$, $dx$, and the limits into $I$:

$I = \int_{\sqrt{2}}^{2} (t^2 - 2) \cdot \sqrt{t^2} \cdot (2t) dt$

$I = \int_{\sqrt{2}}^{2} (t^2 - 2) \cdot t \cdot 2t dt$

$I = 2 \int_{\sqrt{2}}^{2} (t^4 - 2t^2) dt$ [By the Linearity Property of Integrals]

Step 4: Evaluation via Fundamental Theorem of Calculus

Finding the antiderivative:

$I = 2 \left[ \frac{t^5}{5} - \frac{2t^3}{3} \right]_{\sqrt{2}}^{2}$

Evaluating at the boundaries:

$I = 2 \left[ \left( \frac{2^5}{5} - \frac{2(2^3)}{3} \right) - \left( \frac{(\sqrt{2})^5}{5} - \frac{2(\sqrt{2})^3}{3} \right) \right]$

$I = 2 \left[ \left( \frac{32}{5} - \frac{16}{3} \right) - \left( \frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} \right) \right]$

Step 5: Algebraic Simplification

Finding common denominators:

$I = 2 \left[ \left( \frac{96 - 80}{15} \right) - \left( \frac{12\sqrt{2} - 20\sqrt{2}}{15} \right) \right]$

$I = 2 \left[ \frac{16}{15} - \left( \frac{-8\sqrt{2}}{15} \right) \right]$

$I = 2 \left[ \frac{16 + 8\sqrt{2}}{15} \right] = \frac{32 + 16\sqrt{2}}{15}$

Visual Representation of the Area Under Curve

Final Solution: $\frac{16(2 + \sqrt{2})}{15}$

Step 1: Formal Definition of the Definite Integral

We are tasked with evaluating the following definite integral:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \cos^2 x} \, dx$

Step 2: Selection of Substitution Variable

To simplify the integrand, we utilize the method of substitution [Per the Substitution Rule for Integrals]. Let:

$u = \cos x$

Differentiating $u$ with respect to $x$ gives:

$\frac{du}{dx} = -\sin x \implies du = -\sin x \, dx \implies -du = \sin x \, dx$

Step 3: Transformation of Limits of Integration

We must map the limits of integration from the $x$-domain to the $u$-domain:

• When $x = 0$: $u = \cos(0) = 1$
• When $x = \frac{\pi}{2}$: $u = \cos(\frac{\pi}{2}) = 0$

Step 4: Substitution and Algebraic Manipulation

Substituting the variables and limits into the integral expression:

$I = \int_{1}^{0} \frac{-du}{1 + u^2}$

By the property of definite integrals, $\int_{a}^{b} f(u) \, du = -\int_{b}^{a} f(u) \, du$, we can eliminate the negative sign by swapping the limits [Per the Fundamental Theorem of Calculus]:

$I = \int_{0}^{1} \frac{1}{1 + u^2} \, du$

Step 5: Applying Standard Integral Formula

The integrand follows the standard form $\int \frac{1}{1+u^2} \, du = \arctan(u) + C$. Applying the Fundamental Theorem of Calculus [Per the Second Fundamental Theorem of Calculus]:

$I = \left[ \arctan(u) \right]_{0}^{1}$

Evaluating at the boundaries:

$I = \arctan(1) - \arctan(0)$

Step 6: Numerical Calculation

Utilizing the values of inverse trigonometric functions:

• $\arctan(1) = \frac{\pi}{4}$
• $\arctan(0) = 0$

$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$

Final Solution: \frac{\pi}{4}

Step 1: Completing the Square for the Quadratic Denominator

To evaluate the integral $I = \int_{0}^{2} \frac{dx}{-x^2 + x + 4}$, we first express the quadratic expression in the denominator as a sum or difference of squares.

Factor out the negative sign from the quadratic expression: $-x^2 + x + 4 = -(x^2 - x - 4)$

To complete the square for $x^2 - x$, we take half of the coefficient of $x$ (which is $-1/2$), square it ($1/4$), and add/subtract it inside the parenthesis:

$x^2 - x - 4 = (x^2 - x + \frac{1}{4}) - \frac{1}{4} - 4$

$= (x - \frac{1}{2})^2 - \frac{17}{4}$

Substituting this back into the original expression: $-(x^2 - x - 4) = \frac{17}{4} - (x - \frac{1}{2})^2$

Step 2: Setting up the Integral in Standard Form

The integral now takes the form: $I = \int_{0}^{2} \frac{dx}{(\frac{\sqrt{17}}{2})^2 - (x - \frac{1}{2})^2}$

We utilize the standard integral formula: $\int \frac{dx}{a^2 - u^2} = \frac{1}{2a} \ln \left| \frac{a+u}{a-u} \right| + C$ [Per the table of Standard Integrals].

Here, $a = \frac{\sqrt{17}}{2}$ and $u = (x - \frac{1}{2})$.

Step 3: Applying Integration and Evaluating Limits

Applying the formula: $I = \left[ \frac{1}{2(\frac{\sqrt{17}}{2})} \ln \left| \frac{\frac{\sqrt{17}}{2} + (x - \frac{1}{2})}{\frac{\sqrt{17}}{2} - (x - \frac{1}{2})} \right| \right]_{0}^{2}$

$I = \frac{1}{\sqrt{17}} \left[ \ln \left| \frac{\sqrt{17} + 2x - 1}{\sqrt{17} - 2x + 1} \right| \right]_{0}^{2}$

Evaluating at the upper limit ($x=2$): $\text{Upper} = \ln \left| \frac{\sqrt{17} + 4 - 1}{\sqrt{17} - 4 + 1} \right| = \ln \left| \frac{\sqrt{17} + 3}{\sqrt{17} - 3} \right|$

Evaluating at the lower limit ($x=0$): $\text{Lower} = \ln \left| \frac{\sqrt{17} - 1}{\sqrt{17} + 1} \right|$

Step 4: Final Algebraic Simplification

$I = \frac{1}{\sqrt{17}} \left( \ln \frac{\sqrt{17} + 3}{\sqrt{17} - 3} - \ln \frac{\sqrt{17} - 1}{\sqrt{17} + 1} \right)$

Using the logarithm property $\ln(a) - \ln(b) = \ln(a/b)$: $I = \frac{1}{\sqrt{17}} \ln \left( \frac{\sqrt{17} + 3}{\sqrt{17} - 3} \cdot \frac{\sqrt{17} + 1}{\sqrt{17} - 1} \right)$

$I = \frac{1}{\sqrt{17}} \ln \left( \frac{17 + \sqrt{17} + 3\sqrt{17} + 3}{17 - \sqrt{17} - 3\sqrt{17} + 3} \right) = \frac{1}{\sqrt{17}} \ln \left( \frac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}} \right)$

Dividing numerator and denominator by 4: $I = \frac{1}{\sqrt{17}} \ln \left( \frac{5 + \sqrt{17}}{5 - \sqrt{17}} \right)$

Rationalizing the argument: $\frac{(5+\sqrt{17})^2}{25-17} = \frac{25+17+10\sqrt{17}}{8} = \frac{42+10\sqrt{17}}{8} = \frac{21+5\sqrt{17}}{4}$.

Final Solution: $\frac{1}{\sqrt{17}} \ln \left( \frac{5 + \sqrt{17}}{5 - \sqrt{17}} \right)$ or $\frac{1}{\sqrt{17}} \ln \left( \frac{21+5\sqrt{17}}{4} \right)$

Step 1: Identifying the Integral and Substitution Variable

We are tasked to evaluate the definite integral:

$I = \int_{0}^{1} \frac{x}{x^2+1} dx$

Observe that the derivative of the denominator $x^2 + 1$ is $2x$. Since the numerator contains a factor of $x$, we utilize the Method of Substitution [Per the Substitution Rule for Definite Integrals].

Let $u = x^2 + 1$.

Step 2: Differentiating and Adjusting Differentials

Differentiating $u$ with respect to $x$:

$\frac{du}{dx} = 2x \implies du = 2x \, dx \implies \frac{1}{2} du = x \, dx$

Step 3: Transforming the Limits of Integration

Since the variable of integration changes from $x$ to $u$, we must update the limits of integration accordingly:

• When $x = 0$, $u = (0)^2 + 1 = 1$.
• When $x = 1$, $u = (1)^2 + 1 = 2$.

Step 4: Setting up the Transformed Integral

Substitute $u$ and $du$ into the original integral:

$I = \int_{1}^{2} \frac{1}{u} \cdot \left(\frac{1}{2}\right) du$

$I = \frac{1}{2} \int_{1}^{2} \frac{1}{u} du$

Step 5: Evaluating the Antiderivative

The antiderivative of $\frac{1}{u}$ is $\ln|u|$ [By the Fundamental Theorem of Calculus, Part II]:

$I = \frac{1}{2} [\ln|u|]_{1}^{2}$

$I = \frac{1}{2} (\ln|2| - \ln|1|)$

Step 6: Final Calculation

Knowing that $\ln(1) = 0$:

$I = \frac{1}{2} (\ln 2 - 0)$

$I = \frac{1}{2} \ln 2$

Final Solution: $\frac{1}{2} \ln 2$

Step 1: Completing the Square in the Denominator

The given integral is $I = \int_{-1}^{1} \frac{dx}{x^2+2x+5}$. To simplify the quadratic expression in the denominator, we complete the square [Algebraic Identity: $x^2 + 2ax + a^2 = (x+a)^2$]:

$x^2 + 2x + 5 = (x^2 + 2x + 1) + 4$

$x^2 + 2x + 5 = (x+1)^2 + 2^2$

Thus, the integral becomes: $I = \int_{-1}^{1} \frac{dx}{(x+1)^2 + 2^2}$

Step 2: Applying the Integration Formula

We utilize the standard integral formula for inverse trigonometric functions [Standard Integral Rule]:

$\int \frac{dx}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$

Here, $u = x+1$, $du = dx$, and $a = 2$.

$I = \left[ \frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) \right]_{-1}^{1}$

Step 3: Evaluating the Definite Integral

Applying the Fundamental Theorem of Calculus [FTC Part II]:

$I = \frac{1}{2} \left[ \tan^{-1}\left(\frac{1+1}{2}\right) - \tan^{-1}\left(\frac{-1+1}{2}\right) \right]$

$I = \frac{1}{2} \left[ \tan^{-1}\left(\frac{2}{2}\right) - \tan^{-1}\left(\frac{0}{2}\right) \right]$

$I = \frac{1}{2} \left[ \tan^{-1}(1) - \tan^{-1}(0) \right]$

Step 4: Simplification and Final Result

Substituting the known trigonometric values [Trigonometric Domain values]:

$\tan^{-1}(1) = \frac{\pi}{4}$

$\tan^{-1}(0) = 0$

$I = \frac{1}{2} \left[ \frac{\pi}{4} - 0 \right] = \frac{\pi}{8}$

Visual Representation of the Function

Final Solution: \frac{\pi}{8}

Initial Setup and Theoretical Foundation

To evaluate the definite integral $I = \int_{1}^{2} (\frac{1}{x} - \frac{1}{2x^2}) e^{2x} dx$, we identify the integrand as a structure compatible with the theorem: $\int e^{kx} [f(x) + \frac{f'(x)}{k}] dx = \frac{1}{k} e^{kx} f(x) + C$.

Step 1: Algebraic Manipulation of the Integrand

Let the integrand be $g(x) = (\frac{1}{x} - \frac{1}{2x^2}) e^{2x}$. We rewrite the expression to fit the form $e^{ax} [f(x) + f'(x)]$: Since the exponential term contains $2x$, we aim for a derivative involving $2$. Rewrite: $g(x) = e^{2x} (\frac{1}{x} - \frac{1}{2x^2})$.

Let $f(x) = \frac{1}{2x}$. Then, $f'(x) = \frac{d}{dx}(\frac{1}{2}x^{-1}) = -\frac{1}{2}x^{-2} = -\frac{1}{2x^2}$. [By Power Rule of Differentiation]

Thus, the integrand is of the form $e^{2x} [f'(x) + 2f(x)]$. However, a more direct substitution approach is preferred for absolute clarity.

Step 2: Method of Substitution

Let $u = 2x$. Then $du = 2 dx$, which implies $dx = \frac{du}{2}$. Change of limits: When $x=1, u=2$. When $x=2, u=4$.

Substitute into the integral: $I = \int_{2}^{4} (\frac{1}{u/2} - \frac{1}{2(u/2)^2}) e^u \frac{du}{2}$ $I = \int_{2}^{4} (\frac{2}{u} - \frac{2}{u^2}) e^u \frac{du}{2}$ $I = \int_{2}^{4} (\frac{1}{u} - \frac{1}{u^2}) e^u du$

Step 3: Decomposition and Integration

We use the identity $\int e^u [f(u) + f'(u)] du = e^u f(u) + C$, where $f(u) = \frac{1}{u}$ and $f'(u) = -\frac{1}{u^2}$. $I = \int_{2}^{4} e^u (\frac{1}{u}) du + \int_{2}^{4} e^u (-\frac{1}{u^2}) du$ Using Integration by Parts on $\int \frac{e^u}{u} du$: $\int \frac{e^u}{u} du = \frac{1}{u} e^u - \int e^u (-\frac{1}{u^2}) du$ Therefore, $\int (\frac{1}{u} - \frac{1}{u^2}) e^u du = [\frac{e^u}{u}]_{2}^{4}$

Step 4: Evaluating Definite Limits

$I = [\frac{e^u}{u}]_{2}^{4}$ $I = (\frac{e^4}{4}) - (\frac{e^2}{2})$ $I = \frac{e^4 - 2e^2}{4}$

Final Solution: \frac{e^4 - 2e^2}{4}

Step 1: Analyzing the Integral and Substitution Setup

Given the definite integral:

$I = \int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos^5 \phi \, d\phi$

To simplify, we express $\cos^5 \phi$ as $\cos^4 \phi \cdot \cos \phi$:

$I = \int_{0}^{\frac{\pi}{2}} (\sin \phi)^{1/2} (\cos^2 \phi)^2 \cos \phi \, d\phi$

Applying the Pythagorean identity $\cos^2 \phi = 1 - \sin^2 \phi$ [Per Trigonometric Identity]:

$I = \int_{0}^{\frac{\pi}{2}} (\sin \phi)^{1/2} (1 - \sin^2 \phi)^2 \cos \phi \, d\phi$

Step 2: Performing the Change of Variables

Let $u = \sin \phi$. Then, the derivative is $du = \cos \phi \, d\phi$ [By Differential Calculus].

Adjusting the limits of integration:

• When $\phi = 0$, $u = \sin(0) = 0$.
• When $\phi = \frac{\pi}{2}$, $u = \sin(\frac{\pi}{2}) = 1$.

Step 3: Expanding and Integrating

Substituting $u$ and $du$ into the integral:

$I = \int_{0}^{1} u^{1/2} (1 - u^2)^2 \, du$

Expanding the binomial $(1 - u^2)^2 = 1 - 2u^2 + u^4$ [By Binomial Theorem]:

$I = \int_{0}^{1} u^{1/2} (1 - 2u^2 + u^4) \, du$

$I = \int_{0}^{1} (u^{1/2} - 2u^{5/2} + u^{9/2}) \, du$

Step 4: Applying the Power Rule for Integration

Integrating term by term using $\int x^n dx = \frac{x^{n+1}}{n+1}$ [Per Fundamental Theorem of Calculus]:

$I = \left[ \frac{u^{3/2}}{3/2} - 2 \frac{u^{7/2}}{7/2} + \frac{u^{11/2}}{11/2} \right]_{0}^{1}$

$I = \left[ \frac{2}{3}u^{3/2} - \frac{4}{7}u^{7/2} + \frac{2}{11}u^{11/2} \right]_{0}^{1}$

Step 5: Evaluating the Definite Integral

Evaluating at the boundaries $u=1$ and $u=0$:

$I = (\frac{2}{3}(1) - \frac{4}{7}(1) + \frac{2}{11}(1)) - (0 - 0 + 0)$

$I = \frac{2}{3} - \frac{4}{7} + \frac{2}{11}$

Finding a common denominator ($3 \times 7 \times 11 = 231$):

$I = \frac{2(77) - 4(33) + 2(21)}{231} = \frac{154 - 132 + 42}{231} = \frac{64}{231}$

Final Solution: \frac{64}{231}

Step 1: Setting up the Integral

We are tasked with evaluating the definite integral:

$I = \int_{\frac{1}{3}}^{1} \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx$

Step 2: Algebraic Manipulation of the Integrand

To simplify the integrand, factor out $x^3$ from the term inside the cube root:

$I = \int_{\frac{1}{3}}^{1} \frac{[x^3(x^{-2}-1)]^{\frac{1}{3}}}{x^4} dx$ [Factoring $x^3$ from the numerator expression]
$I = \int_{\frac{1}{3}}^{1} \frac{x(x^{-2}-1)^{\frac{1}{3}}}{x^4} dx$ [Using exponent rule $(ab)^n = a^n b^n$]
$I = \int_{\frac{1}{3}}^{1} \frac{(x^{-2}-1)^{\frac{1}{3}}}{x^3} dx$

Step 3: Substitution Method

Let $u = x^{-2} - 1$. Then, calculate the derivative with respect to $x$:

$\frac{du}{dx} = -2x^{-3}$
$du = -2x^{-3} dx \Rightarrow -\frac{1}{2} du = \frac{1}{x^3} dx$ [Method of Change of Variables]

Step 4: Adjusting Limits of Integration

Transform the limits of integration from $x$ to $u$:

• When $x = \frac{1}{3}$, $u = (\frac{1}{3})^{-2} - 1 = 9 - 1 = 8$.
• When $x = 1$, $u = (1)^{-2} - 1 = 1 - 1 = 0$.

Step 5: Evaluating the Transformed Integral

Substitute the variables and limits into the integral expression:

$I = \int_{8}^{0} u^{\frac{1}{3}} (-\frac{1}{2}) du$
$I = -\frac{1}{2} \int_{8}^{0} u^{\frac{1}{3}} du$
$I = \frac{1}{2} \int_{0}^{8} u^{\frac{1}{3}} du$ [By swapping limits and changing sign of the integral]
$I = \frac{1}{2} \left[ \frac{u^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} \right]_{0}^{8}$ [Using Power Rule $\int x^n dx = \frac{x^{n+1}}{n+1}$]
$I = \frac{1}{2} \left[ \frac{u^{\frac{4}{3}}}{\frac{4}{3}} \right]_{0}^{8}$
$I = \frac{1}{2} \cdot \frac{3}{4} [u^{\frac{4}{3}}]_{0}^{8}$
$I = \frac{3}{8} [8^{\frac{4}{3}} - 0]$
$I = \frac{3}{8} [(2^3)^{\frac{4}{3}}]$
$I = \frac{3}{8} [2^4] = \frac{3}{8} \cdot 16 = 3 \cdot 2 = 6$

Final Solution: 6

Step 1: Application of the Leibniz Integral Rule

To find the derivative of a function defined as a definite integral with variable limits, we employ the Leibniz Integral Rule [Fundamental Theorem of Calculus, Part 1]. For a function defined as:

$f(x) = \int_{g(x)}^{h(x)} \phi(t) dt$

The derivative $f'(x)$ is given by:

$f'(x) = \phi(h(x)) \cdot h'(x) - \phi(g(x)) \cdot g'(x)$

Step 2: Identification of Components

Given the function:

$f(x) = \int_{0}^{x} t \sin t \, dt$

• $\phi(t) = t \sin t$
• $h(x) = x$, therefore $h'(x) = \frac{d}{dx}(x) = 1$
• $g(x) = 0$, therefore $g'(x) = \frac{d}{dx}(0) = 0$

Step 3: Execution of the Differentiation

Applying the formula derived in Step 1:

$f'(x) = \phi(x) \cdot (1) - \phi(0) \cdot (0)$

$f'(x) = (x \sin x) \cdot 1 - 0$

$f'(x) = x \sin x$

Visual Representation of the Integrand

Step 4: Verification via Integration (Alternative Method)

If we evaluate the integral first using Integration by Parts ($\int u dv = uv - \int v du$):

Let $u = t$, $dv = \sin t \, dt \implies du = dt$, $v = -\cos t$

$\int t \sin t \, dt = -t \cos t - \int (-\cos t) \, dt = -t \cos t + \sin t$

Evaluating from $0$ to $x$:

$f(x) = [-t \cos t + \sin t]_{0}^{x} = (-x \cos x + \sin x) - (-0 \cos 0 + \sin 0) = \sin x - x \cos x$

Now, differentiate $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sin x - x \cos x)$

$f'(x) = \cos x - [x(-\sin x) + \cos x(1)]$

$f'(x) = \cos x + x \sin x - \cos x$

$f'(x) = x \sin x$

Final Solution: $f'(x) = x \sin x$