Step 1: Formal Definition of the Integral

Let the given definite integral be defined as $I$:

$I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \, dx$ --- (Equation 1)

Step 2: Applying the King's Property of Definite Integrals

We utilize the property: $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$ [Per the King's Property of Definite Integrals].

Substituting $x$ with $(a-x)$ in Equation 1:

$I = \int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{a-(a-x)}} \, dx$

$I = \int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} \, dx$ --- (Equation 2)

Step 3: Summation of Equations

Adding Equation 1 and Equation 2:

$I + I = \int_{0}^{a} \left( \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} + \frac{\sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}} \right) \, dx$

$2I = \int_{0}^{a} \left( \frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}} \right) \, dx$

Step 4: Algebraic Simplification and Evaluation

Since $\frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}} = 1$ for all $x \in (0, a)$:

$2I = \int_{0}^{a} 1 \, dx$

$2I = [x]_{0}^{a}$ [By the Fundamental Theorem of Calculus]

$2I = a - 0$

$2I = a$

Step 5: Solving for I

$I = \frac{a}{2}$

Visual Representation of the Area Under the Curve

Final Solution: \frac{a}{2}

Step 1: Analyzing the Integrand and Property Application

The integral to evaluate is $I = \int_{0}^{4} |x-1| \, dx$. The integrand involves an absolute value function $|f(x)|$, which changes its definition at the point where the argument is zero. Specifically, $|x-1| = 0$ at $x = 1$.

We apply the property of definite integrals: $\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx$ where $a < c < b$. [Property: Additivity of Interval of Integration]

Since the interval is $[0, 4]$, we split the integral at $c = 1$:

$I = \int_{0}^{1} |x-1| \, dx + \int_{1}^{4} |x-1| \, dx$

Step 2: Defining the Piecewise Function

Based on the definition of absolute value:

• For $x \in [0, 1]$, $x-1 \le 0$, thus $|x-1| = -(x-1) = 1-x$.
• For $x \in [1, 4]$, $x-1 \ge 0$, thus $|x-1| = x-1$.

Substituting these into the split integrals:

$I = \int_{0}^{1} (1-x) \, dx + \int_{1}^{4} (x-1) \, dx$

Step 3: Integration and Evaluation

Applying the Fundamental Theorem of Calculus: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$.

For the first part:

$\int_{0}^{1} (1-x) \, dx = \left[ x - \frac{x^2}{2} \right]_{0}^{1} = \left( 1 - \frac{1^2}{2} \right) - (0 - 0) = 1 - 0.5 = 0.5$

For the second part:

$\int_{1}^{4} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{4} = \left( \frac{4^2}{2} - 4 \right) - \left( \frac{1^2}{2} - 1 \right)$ $= (8 - 4) - (0.5 - 1) = 4 - (-0.5) = 4 + 0.5 = 4.5$

Step 4: Summation

$I = 0.5 + 4.5 = 5.0$

Visual Representation of the Area

Final Solution: The value of the integral $\int_{0}^{4} |x-1| \, dx$ is $5$.

Given Integral and Property Setup

The objective is to evaluate the definite integral:

$I = \int_{2}^{8} |x-5| dx$

[By the property of definite integrals: $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$, where $a < c < b$]

Step 1: Analyzing the Modulus Function

The integrand $f(x) = |x-5|$ changes its definition based on the value of $x$ relative to the critical point $x = 5$:

• $|x-5| = -(x-5)$ for $x < 5$
• $|x-5| = (x-5)$ for $x \ge 5$

Since the interval of integration is $[2, 8]$ and the point $5$ lies within this interval, we split the integral at $x = 5$:

$I = \int_{2}^{5} -(x-5) dx + \int_{5}^{8} (x-5) dx$

Step 2: Evaluating the Split Integrals

Part A: Evaluating $\int_{2}^{5} -(x-5) dx$

$= -\left[ \frac{x^2}{2} - 5x \right]_{2}^{5}$

$= -\left[ (\frac{25}{2} - 25) - (\frac{4}{2} - 10) \right]$

$= -\left[ -12.5 - (-8) \right] = -(-4.5) = 4.5$

Part B: Evaluating $\int_{5}^{8} (x-5) dx$

$= \left[ \frac{x^2}{2} - 5x \right]_{5}^{8}$

$= \left[ (\frac{64}{2} - 40) - (\frac{25}{2} - 25) \right]$

$= \left[ (32 - 40) - (-12.5) \right] = -8 + 12.5 = 4.5$

Step 3: Calculating Total Sum

$I = 4.5 + 4.5 = 9$

Geometric Representation

Final Solution: 9

Step 1: Simplifying the Integrand

The given integral is $I = \int_{0}^{\frac{\pi}{2}} (2\log \sin x - \log \sin 2x) \, dx$.

Using the property of logarithms $n\log a = \log a^n$, we can rewrite the first term. Using the double angle formula for sine, $\sin 2x = 2 \sin x \cos x$, we expand the second term:

$I = \int_{0}^{\frac{\pi}{2}} (\log \sin^2 x - \log(2 \sin x \cos x)) \, dx$

Applying the quotient rule for logarithms, $\log a - \log b = \log(\frac{a}{b})$:

$I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{\sin^2 x}{2 \sin x \cos x}\right) \, dx$

$I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{\sin x}{2 \cos x}\right) \, dx$

$I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2} \tan x\right) \, dx$

Step 2: Decomposition using Logarithmic Properties

Using the property $\log(ab) = \log a + \log b$:

$I = \int_{0}^{\frac{\pi}{2}} (\log(2^{-1}) + \log \tan x) \, dx$

$I = \int_{0}^{\frac{\pi}{2}} -\log 2 \, dx + \int_{0}^{\frac{\pi}{2}} \log \tan x \, dx$

Step 3: Evaluating the Separate Integrals

Part A: $\int_{0}^{\frac{\pi}{2}} -\log 2 \, dx$

$= -\log 2 [x]_{0}^{\frac{\pi}{2}} = -\log 2 (\frac{\pi}{2} - 0) = -\frac{\pi}{2} \log 2$

Part B: $\int_{0}^{\frac{\pi}{2}} \log \tan x \, dx$

Let $J = \int_{0}^{\frac{\pi}{2}} \log \tan x \, dx$. By the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ [King's Property]:

$J = \int_{0}^{\frac{\pi}{2}} \log \tan(\frac{\pi}{2} - x) \, dx = \int_{0}^{\frac{\pi}{2}} \log \cot x \, dx$

$J = \int_{0}^{\frac{\pi}{2}} \log(\frac{1}{\tan x}) \, dx = \int_{0}^{\frac{\pi}{2}} -\log \tan x \, dx = -J$

Therefore, $2J = 0 \implies J = 0$.

Step 4: Final Summation

$I = (-\frac{\pi}{2} \log 2) + 0$

$I = -\frac{\pi}{2} \log 2$

Final Solution: -\frac{\pi}{2} \log 2

Step 1: Definition of the Definite Integral

Let the given integral be $I$:

$I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\sin x}{4+3\cos x} \right) dx$ [Equation 1]

Step 2: Application of King's Property

Utilize the definite integral property: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. Here, $a=0$ and $b=\frac{\pi}{2}$.

$I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\sin(\frac{\pi}{2}-x)}{4+3\cos(\frac{\pi}{2}-x)} \right) dx$

Applying trigonometric co-function identities $\sin(\frac{\pi}{2}-x) = \cos x$ and $\cos(\frac{\pi}{2}-x) = \sin x$ [Per Trigonometric Identity Axioms]:

$I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\cos x}{4+3\sin x} \right) dx$ [Equation 2]

Step 3: Summation of Equations

Add Equation 1 and Equation 2:

$2I = \int_{0}^{\frac{\pi}{2}} \left[ \log \left( \frac{4+3\sin x}{4+3\cos x} \right) + \log \left( \frac{4+3\cos x}{4+3\sin x} \right) \right] dx$

Employ the logarithmic property $\log(a) + \log(b) = \log(ab)$ [Per Logarithmic Laws]:

$2I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\sin x}{4+3\cos x} \cdot \frac{4+3\cos x}{4+3\sin x} \right) dx$

Step 4: Algebraic Simplification

The terms within the logarithm cancel out to yield unity:

$2I = \int_{0}^{\frac{\pi}{2}} \log(1) dx$

Since $\log(1) = 0$ [By logarithmic definition]:

$2I = \int_{0}^{\frac{\pi}{2}} 0 \cdot dx$ $2I = 0$ $I = 0$

Visual Representation of the Integrand

Final Solution: 0

Step 1: Identifying the Property for Definite Integration

To evaluate the integral $I = \int_{0}^{2} x \sqrt{2-x} \, dx$, we employ the King's Property of definite integrals [Definite Integral Property P4]:

$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$

Here, $a = 0$ and $b = 2$, so we substitute $x$ with $(0+2-x)$, which is $(2-x)$.

Step 2: Applying the Substitution Property

Applying the transformation $x \to (2-x)$ to the original integral:

$I = \int_{0}^{2} (2-x) \sqrt{2-(2-x)} \, dx$

Simplifying the radicand:

$I = \int_{0}^{2} (2-x) \sqrt{x} \, dx$ [Algebraic simplification]

Step 3: Algebraic Expansion

Distribute $\sqrt{x}$ (or $x^{1/2}$) across the binomial term $(2-x)$:

$I = \int_{0}^{2} (2x^{1/2} - x \cdot x^{1/2}) \, dx$

$I = \int_{0}^{2} (2x^{1/2} - x^{3/2}) \, dx$ [Law of Exponents: $x^m \cdot x^n = x^{m+n}$]

Step 4: Evaluating the Definite Integral

Integrate term-by-term using the Power Rule for Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$:

$I = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} \right]_{0}^{2}$

$I = \left[ \frac{4}{3} x^{3/2} - \frac{2}{5} x^{5/2} \right]_{0}^{2}$

Step 5: Applying Limits of Integration

Substitute the upper limit ($x=2$) and lower limit ($x=0$):

$I = \left( \frac{4}{3} (2)^{3/2} - \frac{2}{5} (2)^{5/2} \right) - (0 - 0)$

Noting that $2^{3/2} = 2\sqrt{2}$ and $2^{5/2} = 4\sqrt{2}$:

$I = \frac{4}{3} (2\sqrt{2}) - \frac{2}{5} (4\sqrt{2})$

$I = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5}$

Step 6: Final Arithmetic Simplification

Find a common denominator to combine the terms:

$I = 8\sqrt{2} \left( \frac{1}{3} - \frac{1}{5} \right)$

$I = 8\sqrt{2} \left( \frac{5-3}{15} \right)$

$I = 8\sqrt{2} \left( \frac{2}{15} \right)$

$I = \frac{16\sqrt{2}}{15}$

Geometric Representation

Final Solution: \frac{16\sqrt{2}}{15}

Given Variables & Initial Setup

Let the given integral be denoted by $I$:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx$ — (Equation 1)

Step 1: Applying the Definite Integral Property

We utilize the property of definite integrals, specifically King's Property: $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx$. [Per Property of Definite Integrals, P4]

Here, $a = 0$ and $b = \frac{\pi}{2}$. Therefore, we replace $x$ with $(\frac{\pi}{2} - x)$:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 (\frac{\pi}{2} - x)}{\sin^5 (\frac{\pi}{2} - x) + \cos^5 (\frac{\pi}{2} - x)} \, dx$

Step 2: Trigonometric Reduction

Using the co-function identities $\cos(\frac{\pi}{2} - x) = \sin x$ and $\sin(\frac{\pi}{2} - x) = \cos x$:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x}{\cos^5 x + \sin^5 x} \, dx$ — (Equation 2)

Step 3: Summation of Equations

Adding Equation 1 and Equation 2:

$I + I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx + \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x}{\sin^5 x + \cos^5 x} \, dx$

$2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 x + \sin^5 x}{\sin^5 x + \cos^5 x} \, dx$

Step 4: Simplification and Evaluation

The integrand simplifies to $1$:

$2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx$

$2I = [x]_{0}^{\frac{\pi}{2}}$ [By the Fundamental Theorem of Calculus]

$2I = \frac{\pi}{2} - 0$

$2I = \frac{\pi}{2}$

Step 5: Solving for I

Dividing both sides by 2:

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

Visual Representation of the Integrand

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

Step 1: Analyzing the Integrand

Let the given integral be denoted by $I$:

$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx$

Define the function $f(x) = \sin^2 x$. We test for parity [Per the property of Even and Odd functions]:

$f(-x) = \sin^2(-x) = (-\sin x)^2 = \sin^2 x = f(x)$

Since $f(-x) = f(x)$, the function $f(x) = \sin^2 x$ is an even function.

Step 2: Applying Symmetry Properties

Using the definite integral property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(x)$ is even:

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

Step 3: Trigonometric Identity Substitution

To integrate $\sin^2 x$, we utilize the double-angle identity [Per the Cosine Double Angle Formula]:

$\cos(2x) = 1 - 2\sin^2 x \implies \sin^2 x = \frac{1 - \cos(2x)}{2}$

Substitute this into the integral:

$I = 2 \int_{0}^{\frac{\pi}{2}} \left( \frac{1 - \cos(2x)}{2} \right) \, dx$

Step 4: Evaluating the Integral

Simplify the constants and integrate term by term:

$I = \int_{0}^{\frac{\pi}{2}} (1 - \cos(2x)) \, dx$

$I = \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{2}}$

Step 5: Applying Fundamental Theorem of Calculus

Evaluate at the upper and lower limits:

$I = \left( \frac{\pi}{2} - \frac{\sin(2 \cdot \frac{\pi}{2})}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right)$

$I = \left( \frac{\pi}{2} - \frac{\sin(\pi)}{2} \right) - (0 - 0)$

Since $\sin(\pi) = 0$:

$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

Visual Representation of the Area

Final Solution: $\frac{\pi}{2}$

Given Integral and Initial Setup

Let the given definite integral be denoted by $I$:

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

Step 1: Applying the Definite Integral Property

We utilize the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$ [King’s Property]. Applying this to our integral where $a = \frac{\pi}{2}$:

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

Using the trigonometric identities $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$:

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

Step 2: Adding the Original and Transformed Integrals

Adding the initial expression for $I$ and the newly derived expression:

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

Step 3: Simplification and Evaluation

Observe that the numerators cancel each other out:

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

Consequently:

$I = 0$

Visual Representation of the Integrand

The function $f(x) = \frac{\sin x - \cos x}{1 + \sin x \cos x}$ exhibits point symmetry about the point $(\frac{\pi}{4}, 0)$ within the interval $[0, \frac{\pi}{2}]$. Specifically, $f(\frac{\pi}{2} - x) = -f(x)$, which dictates that the net area under the curve over the specified symmetric interval must be zero [By definition of odd symmetry about the midpoint].

Final Solution: $0$

Step 1: Application of Definite Integral Properties

We are tasked with evaluating the definite integral:

$I = \int_{0}^{2\pi} \cos^5 x \, dx$

To solve this, we apply the property of definite integrals involving symmetric bounds, specifically the property regarding periodicity: $\int_{0}^{2a} f(x) \, dx$ where $f(2a - x) = f(x)$. [Per the property: $\int_{0}^{2a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx$ if $f(2a-x) = f(x)$, and $0$ if $f(2a-x) = -f(x)$].

Step 2: Testing Symmetry of the Integrand

Let $f(x) = \cos^5 x$. We evaluate $f(2\pi - x)$:

$f(2\pi - x) = \cos^5(2\pi - x)$

Since $\cos(2\pi - \theta) = \cos \theta$ [By the Quadrant Rule for Trigonometric Functions], we have:

$f(2\pi - x) = (\cos x)^5 = \cos^5 x = f(x)$

Thus, $I = 2 \int_{0}^{\pi} \cos^5 x \, dx$.

Step 3: Further Symmetry Decomposition

Now consider the integral $\int_{0}^{\pi} \cos^5 x \, dx$. We test the property $\int_{0}^{a} f(x) \, dx = 0$ if $f(a-x) = -f(x)$.

Let $f(x) = \cos^5 x$ and $a = \pi$:

$f(\pi - x) = \cos^5(\pi - x)$

Since $\cos(\pi - x) = -\cos x$ [By the Supplementary Angle Identity], we have:

$f(\pi - x) = (-\cos x)^5 = -\cos^5 x = -f(x)$

Since $f(\pi - x) = -f(x)$, it follows that $\int_{0}^{\pi} \cos^5 x \, dx = 0$.

Step 4: Logical Conclusion

Substituting the result from Step 3 back into the expression derived in Step 2:

$I = 2 \times 0 = 0$

Visual Representation of the Function $\cos^5 x$

Final Solution: $\int_{0}^{2\pi} \cos^5 x \, dx = 0$

Step 1: Formal Definition of the Integral

Let the given definite integral be defined as $I$:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$ --- (Equation 1)

Step 2: Application of Definite Integral Property

We utilize the King's Property, which states that for any continuous function $f(x)$ on $[a, b]$:

$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx$ [Ref: Property of Definite Integrals P4]

Here, $a = 0$ and $b = \frac{\pi}{2}$. Thus, we substitute $x$ with $(\frac{\pi}{2} - x)$:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin(\frac{\pi}{2} - x)}}{\sqrt{\sin(\frac{\pi}{2} - x)} + \sqrt{\cos(\frac{\pi}{2} - x)}} \, dx$

Step 3: Simplification using Trigonometric Identities

Employing the co-function identities $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$ [Ref: Trigonometric Complementary Angle Theorem]:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx$ --- (Equation 2)

Step 4: Algebraic Summation of Integrals

Add Equation 1 and Equation 2:

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

$2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) \, dx$

$2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx$

Step 5: Final Integration and Evaluation

Evaluating the integral of the constant $1$ with respect to $x$:

$2I = [x]_{0}^{\frac{\pi}{2}}$

$2I = \frac{\pi}{2} - 0$

$2I = \frac{\pi}{2}$

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

Geometric Representation (Integrand Behavior)

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

Step 1: Establishing the Integral Definition

We are tasked with evaluating the definite integral:

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

Step 2: Applying the Definite Integral Property

We utilize the property of definite integrals, specifically King's Property: $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. [Per the Fourth Property of Definite Integrals].

Here, $a = 0$ and $b = \frac{\pi}{2}$, so $a+b-x = \frac{\pi}{2} - x$.

$I = \int_{0}^{\frac{\pi}{2}} \cos^2 \left( \frac{\pi}{2} - x \right) \, dx$

Since $\cos(\frac{\pi}{2} - x) = \sin x$ [By Trigonometric Cofunction Identity], the integral becomes:

$I = \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx$ — (Equation 1)

Step 3: Combining Equations

Adding the original integral $I$ and Equation 1:

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

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

Applying the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ [By Fundamental Trigonometric Identity]:

$2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx$

Step 4: Evaluating the Resultant Integral

$2I = [x]_{0}^{\frac{\pi}{2}}$ [By Fundamental Theorem of Calculus]

$2I = \left( \frac{\pi}{2} - 0 \right)$

$2I = \frac{\pi}{2}$

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

Visual Representation of the Function $y = \cos^2 x$

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

Given Integral and Functional Analysis

The integral to be evaluated is:
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^3 + x \cos x + \tan^5 x + 1) \, dx$

Let $f(x) = x^3 + x \cos x + \tan^5 x + 1$. We decompose this into the sum of functions:
$f_1(x) = x^3$
$f_2(x) = x \cos x$
$f_3(x) = \tan^5 x$
$f_4(x) = 1$

Step 1: Symmetry Property Application

We utilize the property of definite integrals:
$\int_{-a}^{a} f(x) \, dx = \begin{cases} 2 \int_{0}^{a} f(x) \, dx & \text{if } f(x) \text{ is even} \\ 0 & \text{if } f(x) \text{ is odd} \end{cases}$
Where $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$. [Per the Property of Definite Integrals over Symmetric Intervals].

Step 2: Checking Parity of Individual Terms

• For $f_1(x) = x^3$: $f_1(-x) = (-x)^3 = -x^3 = -f_1(x)$. Thus, $f_1(x)$ is odd.
• For $f_2(x) = x \cos x$: $f_2(-x) = (-x) \cos(-x) = -x \cos x = -f_2(x)$ (since $\cos(-x) = \cos x$). Thus, $f_2(x)$ is odd.
• For $f_3(x) = \tan^5 x$: $f_3(-x) = \tan^5(-x) = (-\tan x)^5 = -\tan^5 x = -f_3(x)$. Thus, $f_3(x)$ is odd.
• For $f_4(x) = 1$: $f_4(-x) = 1 = f_4(x)$. Thus, $f_4(x)$ is even.

Step 3: Calculating the Definite Integral

By the linearity of integration:
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^5 x \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx$

Since $f_1, f_2, f_3$ are odd functions, their integrals over $[-\frac{\pi}{2}, \frac{\pi}{2}]$ are $0$.
$I = 0 + 0 + 0 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx$

Integrating $f_4(x) = 1$:
$I = [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$
$I = \frac{\pi}{2} - (-\frac{\pi}{2})$
$I = \frac{\pi}{2} + \frac{\pi}{2} = \pi$

Final Solution: \pi

Step 1: Identifying the Definite Integral and Applying Property

Let the given integral be denoted as $I$:

$I = \int_{0}^{\frac{\pi}{4}} \log(1+\tan x) \, dx$ --- (Equation 1)

We utilize the King's Property of definite integrals, which states that for a function $f(x)$ continuous on $[a, b]$:

$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx$ [By Property of Definite Integrals]

Here, $a = 0$ and $b = \frac{\pi}{4}$. Thus, $a + b - x = \frac{\pi}{4} - x$.

Step 2: Substitution and Simplification

Substituting $x \to \frac{\pi}{4} - x$ into Equation 1:

$I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right) \, dx$

Applying the trigonometric identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:

$\tan\left(\frac{\pi}{4} - x\right) = \frac{\tan(\frac{\pi}{4}) - \tan x}{1 + \tan(\frac{\pi}{4})\tan x} = \frac{1 - \tan x}{1 + \tan x}$ [Since $\tan(\frac{\pi}{4}) = 1$]

Substitute this into the integral:

$I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) \, dx$

Step 3: Algebraic Reduction

Combine the terms inside the logarithm:

$1 + \frac{1 - \tan x}{1 + \tan x} = \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} = \frac{2}{1 + \tan x}$

Therefore, the integral becomes:

$I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan x}\right) \, dx$

Using the logarithmic property $\log(\frac{A}{B}) = \log A - \log B$:

$I = \int_{0}^{\frac{\pi}{4}} (\log 2 - \log(1 + \tan x)) \, dx$

Step 4: Solving for I

Split the integral:

$I = \int_{0}^{\frac{\pi}{4}} \log 2 \, dx - \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx$

Observing that the second term is the original integral $I$ (from Equation 1):

$I = \int_{0}^{\frac{\pi}{4}} \log 2 \, dx - I$

$2I = \log 2 \cdot [x]_{0}^{\frac{\pi}{4}}$

$2I = \log 2 \cdot \left(\frac{\pi}{4} - 0\right)$

$2I = \frac{\pi}{4} \log 2$

$I = \frac{\pi}{8} \log 2$

Final Solution: $\frac{\pi}{8} \log 2$

Step 1: Application of Definite Integral Property

To evaluate the definite integral $I = \int_{0}^{\pi} \frac{x}{1 + \sin x} dx$, we employ the King's Property of definite integrals: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$ [Property P4].

Applying this to our integral where $a=0$ and $b=\pi$: $I = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin(\pi - x)} dx$

Since $\sin(\pi - x) = \sin x$ [Trigonometric Identity], the expression becomes: $I = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin x} dx \quad \dots (1)$

Step 2: Combining Integrals

Adding the original integral $I = \int_{0}^{\pi} \frac{x}{1 + \sin x} dx$ and equation $(1)$: $2I = \int_{0}^{\pi} \frac{x + \pi - x}{1 + \sin x} dx$ $2I = \pi \int_{0}^{\pi} \frac{1}{1 + \sin x} dx$

Step 3: Rationalization and Trigonometric Simplification

Multiply the integrand by $\frac{1 - \sin x}{1 - \sin x}$: $2I = \pi \int_{0}^{\pi} \frac{1 - \sin x}{1 - \sin^2 x} dx$ $2I = \pi \int_{0}^{\pi} \frac{1 - \sin x}{\cos^2 x} dx$ [Per Pythagorean identity $\sin^2 x + \cos^2 x = 1$] $2I = \pi \left[ \int_{0}^{\pi} \sec^2 x \, dx - \int_{0}^{\pi} \sec x \tan x \, dx \right]$

Step 4: Evaluating the Integral

Performing the integration: $2I = \pi \Big[ \tan x - \sec x \Big]_{0}^{\pi}$

Applying the Fundamental Theorem of Calculus: $2I = \pi \Big[ (\tan \pi - \sec \pi) - (\tan 0 - \sec 0) \Big]$ $2I = \pi \Big[ (0 - (-1)) - (0 - 1) \Big]$ $2I = \pi [ 1 + 1 ]$ $2I = 2\pi$

Step 5: Visual Representation of the Function

Final Calculation

Solving for $I$: $I = \frac{2\pi}{2} = \pi$

Final Solution: \pi

Step 1: Establishing the Initial Integral

Let the given integral be denoted by $I$:

$I = \int_{0}^{a} f(x)g(x) \, dx$ --- (Equation 1)

[By the property of definite integrals: $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$]

Applying this property to Equation 1:

$I = \int_{0}^{a} f(a-x)g(a-x) \, dx$

Given the condition $f(x) = f(a-x)$, we substitute $f(a-x)$ with $f(x)$:

$I = \int_{0}^{a} f(x)g(a-x) \, dx$ --- (Equation 2)

Step 2: Combining the Integrals

Adding Equation 1 and Equation 2:

$I + I = \int_{0}^{a} f(x)g(x) \, dx + \int_{0}^{a} f(x)g(a-x) \, dx$

$2I = \int_{0}^{a} f(x) [g(x) + g(a-x)] \, dx$

Step 3: Applying Given Condition and Evaluating

Given that $g(x) + g(a-x) = 4$, we substitute this constant value into the integrand:

$2I = \int_{0}^{a} f(x) \cdot (4) \, dx$

$2I = 4 \int_{0}^{a} f(x) \, dx$

Step 4: Algebraic Simplification

Dividing both sides of the equation by 2:

$I = \frac{4}{2} \int_{0}^{a} f(x) \, dx$

$I = 2 \int_{0}^{a} f(x) \, dx$

Visual Representation of Integral Symmetry

Final Solution: $\int_{0}^{a} f(x) g(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$

Step 1: Formal Definition of the Integral

Let the given integral be denoted as $I$:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$ --- (Equation 1)

[By the property of definite integrals: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$]

Step 2: Application of Definite Integral Property

Applying the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ where $a = \frac{\pi}{2}$:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}}(\frac{\pi}{2} - x)}{\sin^{\frac{3}{2}}(\frac{\pi}{2} - x) + \cos^{\frac{3}{2}}(\frac{\pi}{2} - x)} dx$

Using trigonometric co-function identities $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$ [Per Complementary Angle Identities]:

$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} dx$ --- (Equation 2)

Step 3: Summation of Equations

Adding Equation 1 and Equation 2:

$I + I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx + \int_{0}^{\frac{\pi}{2}} \frac{\cos^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$ $2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$

Step 4: Simplification and Evaluation

Since the integrand simplifies to $1$:

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

Applying the Fundamental Theorem of Calculus:

$2I = [x]_{0}^{\frac{\pi}{2}}$ $2I = \frac{\pi}{2} - 0$ $2I = \frac{\pi}{2}$ $I = \frac{\pi}{4}$

Visual Representation of the Integrand

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

Step 1: Identification of the Integrand Property

To evaluate the definite integral $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^7 x \, dx$, we must first examine the parity of the function $f(x) = \sin^7 x$.

A function $f(x)$ is defined as even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.

Evaluating the function at $-x$:

$f(-x) = \sin^7(-x)$

[By the trigonometric identity $\sin(-\theta) = -\sin(\theta)$]:

$f(-x) = (-\sin x)^7 = -(\sin^7 x)$ $f(-x) = -f(x)$

Since $f(-x) = -f(x)$, the function $f(x) = \sin^7 x$ is an odd function.

Step 2: Application of Definite Integral Property

We utilize the standard property of definite integrals regarding symmetric intervals:

$\int_{-a}^{a} f(x) \, dx = \begin{cases} 2\int_{0}^{a} f(x) \, dx & \text{if } f(x) \text{ is even} \\ 0 & \text{if } f(x) \text{ is odd} \end{cases}$

[Per the property of definite integrals over symmetric intervals].

Step 3: Execution and Evaluation

Given the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, where $a = \frac{\pi}{2}$, and having established that $\sin^7 x$ is an odd function:

$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^7 x \, dx = 0$

Visual Representation of the Odd Function

The graph shows the symmetry of $\sin^7 x$ across the origin. The area accumulated from $[-\frac{\pi}{2}, 0]$ is equal in magnitude but opposite in sign to the area accumulated from $[0, \frac{\pi}{2}]$, leading to a total sum of zero.

Final Solution: 0

Step 1: Defining the Integral and Applying Definite Integral Property

Let the given integral be denoted as $I$:

$I = \int_{0}^{\pi} \log(1+\cos x) dx$ --- (Equation 1)

We apply the property of definite integrals: $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ [Per the King's Property of Definite Integrals].

Substituting $x$ with $(\pi - x)$:

$I = \int_{0}^{\pi} \log(1+\cos(\pi - x)) dx$

Since $\cos(\pi - x) = -\cos x$ [By trigonometric identity of supplementary angles], we get:

$I = \int_{0}^{\pi} \log(1-\cos x) dx$ --- (Equation 2)

Step 2: Combining the Integrals

Adding Equation 1 and Equation 2:

$2I = \int_{0}^{\pi} [\log(1+\cos x) + \log(1-\cos x)] dx$

Using the logarithmic identity $\log A + \log B = \log(AB)$ [Per the Product Rule of Logarithms]:

$2I = \int_{0}^{\pi} \log((1+\cos x)(1-\cos x)) dx$

$2I = \int_{0}^{\pi} \log(1-\cos^2 x) dx$

Since $1-\cos^2 x = \sin^2 x$ [Per the Pythagorean identity], we have:

$2I = \int_{0}^{\pi} \log(\sin^2 x) dx$

$2I = 2 \int_{0}^{\pi} \log(\sin x) dx$ [Per the power rule of logarithms, $\log(a^n) = n \log a$]

$I = \int_{0}^{\pi} \log(\sin x) dx$

Step 3: Symmetry and Final Evaluation

Since $\sin(\pi - x) = \sin x$, we use the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$ [Per the Symmetry Property]:

$I = 2 \int_{0}^{\pi/2} \log(\sin x) dx$

Using the standard integral value $\int_{0}^{\pi/2} \log(\sin x) dx = -\frac{\pi}{2} \log 2$ [Established via Wallace’s product or substitution methods]:

$I = 2 \times \left( -\frac{\pi}{2} \log 2 \right)$

$I = -\pi \log 2$

Visual Representation of the Integrand

Final Solution: -\pi \log 2

Step 1: Identifying the Property of Absolute Value

Given the integral $I = \int_{-5}^{5} |x+2| \, dx$. The integrand is an absolute value function defined as:

$|x+2| = \begin{cases} -(x+2) & \text{if } x < -2 \\ (x+2) & \text{if } x \ge -2 \end{cases}$

[By the definition of the absolute value function $|f(x)|$]

Step 2: Splitting the Interval of Integration

Since the integrand changes its definition at $x = -2$, we split the integral at this point within the given range $[-5, 5]$:

$I = \int_{-5}^{-2} -(x+2) \, dx + \int_{-2}^{5} (x+2) \, dx$

[By the Additivity Property of Definite Integrals: $\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx$ where $a < b < c$]

Step 3: Evaluating the First Integral ($I_1$)

$I_1 = \int_{-5}^{-2} -(x+2) \, dx = - \left[ \frac{x^2}{2} + 2x \right]_{-5}^{-2}$

$I_1 = - \left[ \left( \frac{(-2)^2}{2} + 2(-2) \right) - \left( \frac{(-5)^2}{2} + 2(-5) \right) \right]$

$I_1 = - \left[ (2 - 4) - (\frac{25}{2} - 10) \right] = - \left[ -2 - (\frac{5}{2}) \right] = - \left[ -\frac{9}{2} \right] = \frac{9}{2}$

Step 4: Evaluating the Second Integral ($I_2$)

$I_2 = \int_{-2}^{5} (x+2) \, dx = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{5}$

$I_2 = \left[ \left( \frac{5^2}{2} + 2(5) \right) - \left( \frac{(-2)^2}{2} + 2(-2) \right) \right]$

$I_2 = \left[ (\frac{25}{2} + 10) - (2 - 4) \right] = \left[ \frac{45}{2} - (-2) \right] = \frac{45}{2} + 2 = \frac{49}{2}$

Step 5: Visual Representation of the Area

Step 6: Summing the Components

$I = I_1 + I_2$

$I = \frac{9}{2} + \frac{49}{2} = \frac{58}{2} = 29$

Final Solution: 29