Definite Integrals

. Evaluate from to : Multiply by : .

Use . Evaluate: This value is not or . Hence none of the given choices matches.

Step 1: Use product-to-sum formula for : Step 2: Rewrite the integral: Step 3: Evaluate integrals: Step 4: Sum and multiply by : Step 5: So the value is , which does not match any option. --- Please verify the options or the integral again. ---

Step 1: Consider the function .
The function is odd, because:

Step 2: When we integrate an odd function over an interval symmetric about the origin, i.e. from to , the result is , provided the function is defined over that interval.

Step 3: The function is not defined at and , as it approaches infinity there. Thus, the integral must be treated as an improper integral.

Step 4: However, if we consider the integral in the principal value sense: it equals , since the positive and negative parts of the curve cancel out symmetrically due to the odd nature of .

Final Answer:

Step 1: Rewrite the integral: Step 2: Integrate using the power rule: Step 3: Calculate : Step 4: Substitute: None of the options match this answer, so please verify the question or options. ---

Step 1: Note that is an odd function because is odd and raising to an odd power preserves oddness: Step 2: The integral of an odd function over symmetric limits is zero: Step 3: Therefore,

Step 1: Rewrite the integrand: Step 2: Split the integral: Step 3: Evaluate the first integral: Step 4: For the second integral, substitute , so , limits change from and : Step 5: The integral is a Beta function and equals Step 6: So, Step 7: Therefore, However, none of the options directly match this. --- Check the problem statement or options again? If instead the integral is: or something similar, let me know! ---

Step 1: Integrate : Step 2: Evaluate definite integral: Wait, result is 0, which contradicts option (B). Let's carefully check: So correct answer is (A) 0. --- Correction: Correct Answer: (A)

Use integration by parts: where Then, Calculate each term: and Therefore,
Final answer:

This is a standard integral. We know that: Hence,
Final answer:

Let Make the substitution , so . Note that Thus, Simplify the denominator: So the integrand becomes Therefore, Add the two expressions for : Hence,
Final answer:

This integral represents the area under the curve of a semicircle of radius . Using the formula: Evaluate from 0 to : At , At , Therefore,