Definite Integral

Concept: Calculus - Indefinite Integration by Substitution.

Step 1: Choose an appropriate substitution. To eliminate the square root in the denominator, let . This implies that .

Step 2: Differentiate the substitution. Differentiate implicitly: , which simplifies to . Rewrite the numerator of the integral, , to utilize this: .

Step 3: Substitute and simplify the integral. Substitute into the original integral: . The in the numerator and denominator cancel out perfectly, leaving a simple polynomial integral: .

Step 4: Integrate and substitute back. Integrate with respect to : . Now, substitute back into the expression: . We can write as . So the expression becomes: .

Step 5: Compare coefficients and calculate . Compare our integrated result with the given expression: . By matching coefficients, we find that and . The question asks for the value of . Substitute the values: . $ $

Concept: Calculus - Definite Integration using Substitution (converting to and ).

Step 1: Convert the secant and cosecant terms to sine and cosine. Rewrite the integral using basic trigonometric ratios: .

Step 2: Manipulate the denominator to create a term. To integrate expressions with fractional powers of sine and cosine, we aim to form and . Multiply and divide the denominator by : . This simplifies to: . Since , the integral becomes .

Step 3: Apply integration by substitution. Let . Then, differentiating both sides gives .

Step 4: Update the limits of integration. When , . When , . The new integral is .

Step 5: Integrate and evaluate the definite integral. Using the power rule , we get: . Now, apply the limits: . Simplify the exponents: . Using exponent rules ( ): . $ $

We have,

1. Simplify the Integrand:
We are given the integral: We know that and . Substituting these into the integral: To simplify the denominator, we find a common denominator: Substituting this back into the integral: Simplifying by multiplying by the reciprocal of the denominator: The terms cancel, leaving:

2. Apply the Property of Definite Integrals:
We can use the property of definite integrals: Let: Then, applying the property: Since and , we have:

3. Add the Two Integrals:
Now, add the original integral and the transformed integral: Simplifying:

4. Solve the Integral:
Let , so that . When , , and when , . Substituting: Simplifying: The integral of is , so: Substituting the limits: Using and , we get:

5. Find :
Dividing both sides by 2:

Final Answer:
Therefore, the value of the integral is:

1. Integration by Parts Formula:
The integration by parts formula is:

2. Choose and :
Let: - -

3. Find and :
To differentiate , we get: - To integrate , we get: -

4. Apply the Integration by Parts Formula:
Now, applying the formula for integration by parts: Simplifying:

5. Solve the Remaining Integral:
Now, let's solve the integral: Use the substitution: - Let , so , or . Substituting into the integral: Simplifying: Integrating: Substituting back :

6. Substitute Back into the Main Integral:
Now, substitute the result from step 5 back into the main integral:

7. Evaluate the Definite Integral:
Now, we evaluate the definite integral: Substituting the limits of integration: * When : * When : Therefore:

Final Answer:

Step 1: Use the identity for sec(x):
We are tasked with solving the integral: We begin by using the trigonometric identity . This allows us to rewrite the integral as:

Step 2: Simplify the logarithm:
We now apply the property of logarithms that , so the integral becomes:

Step 3: Use the known value for the integral of :
From known integral results, we have: Therefore:

Step 4: Substitute the result into the original integral:
Substituting this result back into the original integral:

Final Answer:
Therefore, the value of the integral is:

Let's solve the definite integral step by step.

1. Set up the Integral:
We are given the integral:

2. Use Trigonometric Substitution:
Let , so that .
When , , so .
When , , so .

Substituting into the integral, we get:

3. Simplify the Integral:
Using the identity , we get: Simplifying further: This simplifies to:

4. Use Trigonometric Identity:
We know that: Therefore, the integral becomes:

5. Integrate:
The integral of is , and the integral of 1 is . So, we have:

6. Evaluate the Limits:
Evaluating at the limits: We know that: Substituting these values: Simplifying:

7. Relate to :
We are given that the integral equals . Thus: Multiply both sides by 6: Simplifying:

Therefore, the value of is:

Let us analyze the expression inside the modulus: So, the sign of changes at and . On the interval , break into regions: 1. \ 2. So: Now compute both integrals: Now add both:

Concept: Use the basic integration formulas: Also note:
Step 1: {Rewrite the integral.}
Step 2: {Integrate each term separately.} Thus,
Step 3: {Apply the limits from 1 to 4.} At : At :
Step 4: {Subtract the values.} Simplifying according to the given options gives: Hence, the correct option is (D).

Concept: For absolute value functions, Thus the integral must be split at the point where the expression inside the modulus becomes zero.

Step 1: Find where the expression inside modulus becomes zero. So we split the integral:

Step 2: Evaluate the first integral.

Step 3: Evaluate the second integral.

Step 4: Add the results. Thus,

Concept: For definite integrals of the form we can use the property: This symmetry property is very useful for expressions involving and .
Step 1: {Let}
Step 2: {Replace with .}
Step 3: {Add the two integrals.}
Step 4: {Evaluate the integral.}

Concept: Use the trigonometric identity This identity simplifies the integration.
Step 1: {Substitute the identity into the integral.}
Step 2: {Split the integral.}
Step 3: {Evaluate both integrals.}
Step 4: {Final result.} Thus, the correct option is (B).