Evaluate the integrals in Exercises 53–76.
71. ∫(from -π...

Question

Evaluate the integrals in Exercises 53–76.

71. ∫(from -π/2 to π/2) 2cosθ dθ/(1+(sinθ)²)

Accepted Answer

Welcome back everyone. Evaluate the integral from negative pi divided by 6 up to pi divided by 6 of 5 cosine theta divided by 1 plus 4 sin squared theta d theta. A pi, b 3 pi divided by 2, C5 pi divided by 4, and D 5 pi divided by 2. For this problem we're going to use Cu substitution. Let's suppose that u is equal to sine theta. Then the derivative of you with respect to theta is the derivative of sine, which is cosine theta. And we can now show that cosine theta multiplied by d theta is equal to duu if we multiply both sides by d theta. Now let's change the limits of integration. So the lower limit of integration u1 is going to be sin of pi divided by 6, which is negative of sin of pi divided by 6 because sin is an odd function, right, and that would be -1/22, and the upper limit of integration u2 is sin of pi divided by 6, which is 1/22. So we can now rewrite the integral as integral from 1/2 up to 1/2 of We can factor out the constant which is 5. cosine theta d theta becomes d u, and in the denominator we have 1 + 4 u squared, right? So essentially from here we have an integral in the form of integral duu divided by 1 plus a 2 u squared. And its value is equal to one divided by a. Inverse of tangent of a multiplied by u plus c. So let's go ahead and apply this formula for our definite integral. We get 5 multiplied by 1 divided by a. In this case, a is going to be 2 because 4 is 2 squared, right? So we get 5 multiplied by 1/2. Inverse of tangent of a multiplied by u, we know that a is 2, so we get 2 u. And we are not adding a constant of integration because we have a definite integral. We want to evaluate the result between -1/22 and 1/22. Let's call this result i and let's continue. So i is going to be equal to 5 halves, right, because 5 multiplied by 1/22 can be simply written as 5 halves. We can factor out that constant and in parentheses we will evaluate. Inverse of tangent of 2 multiplied by 1/22. Which gives us 1 minus inverse of tangent of 2 multiplied by -1/2. That's -1, right? So we get 5 halves in parentheses, inverse of tangent of 1 is pi divided by 4, and inverse of tangent of -1 is pi divided by 4. So we get 5 halves multiplied by 2 pi divided by 4, which is pi divided by 2, right? So 5 halves multiplied by. Pi divided by 2. Which is equal to 5 pi divided by 4, and this corresponds to the answer of choice c. Thank you for watching.

Evaluate the integrals in Exercises 53–76.71. ∫(from -π/2 to π/2) 2cosθ dθ/(1+(sinθ)²)

Watch next

Evaluate the integrals in Exercises 53–76.
71. ∫(from -π/2 to π/2) 2cosθ dθ/(1+(sinθ)²)