Integration

To solve this integral, we use the identity for : Thus, the integral becomes: Since is positive in the interval , we can remove the absolute value: Now, integrate: Thus, the integral becomes: Substitute the values of and : Thus, the value of the integral is:

The integral of is: where is the constant of integration.

Step 1: Use a trigonometric identity.
We use the identity to rewrite the integrand. Step 2: Integrate the expression.
After integrating, the result is . Step 3: Conclusion.
Thus, the correct answer is (B).

Step 1: Use partial fraction decomposition to simplify the integrand: Step 2: Multiply both sides by to eliminate the denominators: Step 3: Expand both sides: Step 4: Set up a system of equations by equating the coefficients of like powers of : - Coefficient of : - Coefficient of : - Coefficient of : - Constant term: Step 5: Solve the system of equations: From , we get . Substitute into : From , we get . Substitute into : Step 6: The partial fractions are: Step 7: Now, integrate term by term: The integrals are standard: Thus, the result is:

Step 1: Use the substitution . Then, . The limits of integration change as follows: when , , and when , . Step 2: Now, the second integral becomes: Step 3: Combining the two integrals: Step 4: The integrals now cover the interval from 0 to , so: Thus, the result is proved:

We use integration by parts. Let: Then: Step 1: Apply the integration by parts formula: Step 2: Substitute: Step 3: Simplify the second integral: Step 4: Final result: Thus, the integral is:

Step 1: Recognizing the Standard Integral Formula
We apply the standard integral formula: Step 2: Substituting
Given that , we find . Thus, the integral becomes:

Step 1: Proof of Integral Identity
Using the substitution , we get: Changing limits: Since: we conclude: Step 2: Apply to Sine Function
For , let , so : Since : Final Verification:

Step 1: Use Substitution
Let , so that . Hence, we have: Now, substitute into the integral: Simplifying, we get: Step 2: Decompose Using Partial Fractions
We need to decompose the integrand using partial fractions: To solve for , , and , multiply both sides by to get: Expanding the right-hand side: Now, group the terms based on powers of : Equate the coefficients of like terms: - For : , - For : , - For the constant: . Solving this system of equations: 1. gives , 2. gives , 3. Substituting and into : Hence, and . Step 3: Substitute the Values of , , and
Substituting these values into the partial fraction decomposition: Thus, we can rewrite the integral as: Step 4: Break the Integral Into Separate Terms
Now, split the integral into two parts: Step 5: Evaluate Each Integral
1. , 2. (standard result), 3. (standard result). So, the total integral becomes: Step 6: Substitute Back Into the Solution
Finally, substitute back into the result:

Step 1: Use Standard Integral Formula
The standard formula for the given integral is: Step 2: Proof via Partial Fractions
Rewriting: Using partial fractions: Solving for and , we obtain: Step 3: Integrate

Step 1: Apply Definite Integration for Area
The area under the curve is given by the integral: Step 2: Compute the Integral
Evaluating the integral, we get:

Step 1: Use the Power Reduction Formula
Step 2: Substitute into the Integral
Splitting: Step 3: Evaluate Each Integral

Step 1: Use Integration by Parts
Using the formula: let: Step 2: Compute and
Step 3: Apply Integration by Parts