Evaluate the definite integral using the appropriate substitutions. ∫ π 2 π sin x 1 + c o s 2 x d x \int_{\frac{\pi}{2}}^{\pi}\frac{\sin x}{1+cos^2x}dx

we're asked to evaluate the definite integral using the appropriate substitutions, and we're faced with the integral from pi over two to pi of sine x over one plus cosine squared x d x. Now whenever we have a trig function in both the numerator and denominator, usually we're gonna wanna choose to be choose one of those trig functions to be u and make a substitution accordingly. Now, looking at this here, I have this cosine of being squared. So if I choose u to be the cosine of x, what does that make du? Well, it makes it equal to the derivative of that. So negative sine x times d x. So I can see that that's pretty much what I have in my numerator here. I'm just missing a negative. So I can go ahead and multiply by that negative. And if I do that, I need to multiply by another negative so that I'm not actually changing the value of my integral. So now with that substitution, if I have a u and I have du, I can rewrite this as negative integral one over one plus u squared du. Now this is perfectly of the form that will result in an inverse tangent with an a value of one here. So this is ultimately going to be equal to negative inverse tangent of U plus C. Now here I don't really have to worry about my plus C because I know that I have bounds I'm going to apply at the end here, but just keeping it there for fun for now. Now this is negative inverse tangent. I'm going to go ahead and plug U back in. Remember, it's the cosine of X. So this is inverse tangent of cosine X and then bounding this with my original bound since this is now back in terms of X pi over two to pi. So now we just have to plug those bounds in. So this becomes negative inverse tangent of the cosine of pi and then minus negative. So that's really gonna be a positive inverse tangent of cosine Pi over two. Now cosine of Pi over two is going to be zero and cosine of Pi is negative one. So this is negative inverse tangent of negative one plus the inverse tangent of zero. Now the inverse tangent of zero is also zero, so that term goes away. And then negative inverse tangent of negative one, that is going to be negative Pi over four. So I have negative negative Pi over four, which ultimately results in a positive Pi over four. And this gives me my final answer for my original definite integral. Let us know if you have any questions and I'll see you in the next one.

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11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

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Multiple Choice

Evaluate the definite integral using the appropriate substitutions.
$\int_{\frac{π}{2}}^{π} \frac{\sin x}{1 + c o s^{2} x} d x$

$pi over 4$

$- \frac{π}{4}$

$pi over 8$

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Verified step by step guidance

Recognize that the integral involves a trigonometric function and its derivative. Specifically, the numerator (sin(x)) is the derivative of the denominator's inner function (cos(x)). This suggests a substitution method is appropriate.

Let u = cos(x). Then, the derivative of u with respect to x is du/dx = -sin(x), or equivalently, du = -sin(x) dx. This substitution simplifies the integral.

Change the limits of integration to match the substitution. When x = π/2, u = cos(π/2) = 0. When x = π, u = cos(π) = -1. The new limits of integration are from u = 0 to u = -1.

Rewrite the integral in terms of u. The integral becomes ∫(from u=0 to u=-1) -1/(1+u^2) du. The negative sign comes from the substitution du = -sin(x) dx.

Recognize that the integral -1/(1+u^2) is the derivative of the arctangent function. Use the antiderivative formula ∫1/(1+u^2) du = arctan(u) to evaluate the integral, and then apply the new limits of integration.