Definite Integral

Let:

$ \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \cos(-x) = \cos x 1+e^{-x} \frac{e^x+1}{e^x} I \int_{-\pi/2}^{\pi/2} \cos x dx \boxed{1} $.

$ u = \cos^{-1} x dv = dx du = \frac{-1}{\sqrt{1-x^2}} dx v = x \int u \, dv = uv - \int v \, du \int \frac{x}{\sqrt{1-x^2}} dx w = 1-x^2 dw = -2x dx x dx = -\frac{1}{2} dw \cos^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{3} \cos^{-1} \left( -\frac{1}{2} \right) = \frac{2\pi}{3} $

We need to find the value of .

We can rewrite the integrand as:

Now, we try to write the numerator in terms of and to simplify the integral:

We have . Hence .

Splitting the integral into two terms, .

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Let so .

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Therefore, the correct option is (B) .

Then derivatve .

Consider the option A: derivative is

The derivative of (A) is ...

Therefore, the correct option is (A) 2

Let the given integral be denoted by .

We will use the property of definite integrals: Here, and . Applying the property, we replace with :

Now, add equation (1) and equation (2):

Evaluate the integral:

Solve for :

The value of the integral is 2021. This corresponds to option (B).

We are given the definition of an integral as:

where is a positive integer. We need to find the value of .

Step 1: Establish a recurrence relation for .

Consider the sum for :

Combine the integrals:

Factor out :

Use the identity :

Step 2: Evaluate the integral using substitution.

Let . Then .

Change the limits of integration:

• When , .
• When , .

Substitute into the integral:

Evaluate the integral:

Since , , so .

This is the recurrence relation, valid for .

Step 3: Apply the recurrence relation to find .

We need to find . This matches the form with .

Using the relation with :

So, the correct answer is (D):

The given integral is:

We can simplify the expression inside the integral. By using the identity and , the expression becomes:

After simplifying:

The integral of is:

Thus, the correct answer is: (A)

We are given the integral . We will use integration by parts.

Let:

• , so
• , so

Using the formula for integration by parts, , we get:

Now, simplifying the expression:

At this point, we notice that the integral appears again. Therefore, we can isolate the original integral on one side:

Now, substitute the value of the second integral back into the equation:

Thus, the value of the integral is:

The given integral is: To solve this, use the substitution , so that . The integral becomes: This is a standard integral, and its solution is: Substituting back into the result, we get: Therefore, the correct answer is (D).

Let's solve the integral . We use substitution to solve this integral. Let: Differentiating both sides with respect to , we get: This implies: Substituting into the integral, we get: The integral of is , so: Substituting back , we get: Thus, the correct answer is:

The correct answer is (A) : .

We are given the integral:

Let us simplify the integrand:
Use substitution: Let
Then, Also, , so: Thus the integral becomes: Now change limits: - When , - When , So,

Final Answer:

We are asked to evaluate the definite integral as the limit of a sum.

Step 1: Partition the interval
Let the interval be divided into subintervals, each of width

Step 2: Choose sample points
Using right endpoints:

Step 3: Define the Riemann sum

Step 4: Take the limit

Step 5: Compute the integral

Final Answer:

We are tasked with evaluating the integral:

We can use the identity to rewrite as:

Similarly, can be written as:

Substituting these expressions into the integrand, we have:

Simplifying further:

Factoring out a 2 from the numerator:

Now, we can factor the numerator further using the difference of squares identity:

Canceling out the common factor in the numerator and denominator, we have:

Now we can integrate this expression:

Integrating term by term:

The integral of with respect to is , and the integral of a constant ( ) with respect to is the constant times :

Thus, the integral is equal to:

This corresponds to option (C) .

We evaluate the integral:

Step 1: Use substitution
Let When ,
When , We know: But this approach is complicated. Try different substitution.

Step 2: Let us try
Then, , and So: Simplify:

Step 3: Expand and integrate

Final answer:

The correct answer is: (C):

We are tasked with evaluating the integral:

Step 1: Complete the square

We begin by completing the square inside the square root. The expression inside the square root is . To complete the square, we rewrite it as:

Now the integral becomes:

Step 2: Use substitution

We use the substitution , so that . The integral becomes:

Step 3: Use standard integral formula

We recognize that this is a standard integral of the form , where the solution is:

Substituting , we get:

Step 4: Substitute back and simplify

Substitute back into the result to get the final answer:

Conclusion:
The correct answer is (C):

The correct answer is: (A): 4.

We are tasked with evaluating the integral:

Step 1: Break the integral into separate parts

The given integral has two terms: a polynomial and a trigonometric function. We can break the integral into two separate integrals:

Step 2: Evaluate the polynomial integral

The first part is a straightforward polynomial integral:

This is easy to evaluate using the power rule. Evaluating at the limits gives:

Substituting the limits gives a result of .

Step 3: Evaluate the trigonometric integral

The second part involves a trigonometric function. Using substitution, let , then . This transforms the integral into:

We can now apply integration by parts to solve this integral:

Evaluating this from to gives a result of .

Step 4: Add the results

Now, we add the results from both integrals: .

Conclusion:
The correct answer is (A): 4.