Evaluate the integrals in Exercises 69–134. The integrals are ...

Question

Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.

133. ∫ (sin²x) / (1 + sin²x) dx

Accepted Answer

Evaluate the integral of cosine squared x divided by 1 plus sine square x and dx. Now, to solve this, we'll first rewrite our integrams. We have cosine squared x equals 1. Minus sin squared x, this is one of our Pythagorean identities. Now, from here, we have 1 minus sin 2 x divided by 1 + sin 2 x. This is in the integral. Now we can use algebraic manipulation to rewrite this as -1 + 2 divided by 1 plus sin squared x with dx at the end. Now, let's go ahead and separate our integral. We have the integral. Of -1 dx plus the integral of 1 divided by 1 plus sin squared x d x. Now the 2 is going to go in front of the integral. Now we can use this and rewrite it. As X, which is the integral of -1. Plus 2 i plus c where i is the integral of 1 divided by 1 plus sin squared x d x. Now we can solve for I. To solve for I will make use of a substitution. Let's let T equals tangent of X, for example. That means DT equals sequence 2 XDX. Which is equivalent to 1 plus tangent squared x d x based on another of our Pythagorean identities. This is 1 + T2 DX. We can then solve for dx by dividing, giving us dx equals. DT divided by 1 + T2. We can also say sin 2 x is T2 divided by 1 + T2. We can now writes are integral. We will say i equals. The integral of 1 divided by 1 plus t2 divided by 1 + t2 multiplied by dt divided by 1 + t 2 Now we can simplify this. This simplifies to the integral of dt divided by 1 + 2 t 2. Let's go ahead and solve this now. We can make use of the facts. That the integral of dt. Or do you anarchists. Divided by 1 + U2 is the tangent inverse of U + C. Let's apply it to ours. In our case, U is going to be the square root of 2 T. That makes the d u equivalent. 2, the square root of 2 dt. Which makes DT equals DU divided by the square of 2. We can now use our substitution and rewrite our integral. We have the integral Of DU Divided by 1 + U 2 multiplied by 1 divided by the square of 2. Which is just going to be 1 divided by the square root of 2. Tangent inverse of u plus c. Which is 1 divided by the square root of 2. Tangent inverse. Of the square root of 2 t plus c Now, we can substitute T equals tangent T and combine this with our previous integral. We can write our final integral as x + 2 multiplied. By 1 divided by the square root of 2. Tangent inverse. 2 of 2 Tangent. Of X All plus C. Finally, we can simplify this once more. To give us g x. Plus the square of 2. Tangent inverse square of 2. Tangent X + C. This is the final answer to our problem. OK, I hope to help you solve it. Thank you for watching. Goodbye.

Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.133. ∫ (sin²x) / (1 + sin²x) dx

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Evaluate the integrals in Exercises 69–134. The integrals are listed in random order so you need to decide which integration technique to use.
133. ∫ (sin²x) / (1 + sin²x) dx