$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta}{(1+\cos \theta)^{2}} d \theta$

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Evaluate definite integrals using the chain rule

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\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta}{(1+\cos \theta)^{2}} d \theta

Rewrite using identity: Use the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$ to rewrite the integral. $\int_{0}^{\frac{\pi}{2}}\frac{1 - \cos^2(\theta)}{(1+\cos(\theta))^2}d \theta$
Split into two integrals: Split the integral into two separate integrals. $\int_{0}^{\frac{\pi}{2}}\frac{1}{(1+\cos(\theta))^2}\,d\theta - \int_{0}^{\frac{\pi}{2}}\frac{\cos^2(\theta)}{(1+\cos(\theta))^2}\,d\theta$
Substitute $u$ for $\theta$ : For the first integral, use the substitution $u = 1 + \cos(\theta)$ , $du = -\sin(\theta)d\theta$ . When $\theta = 0$ , $u = 2$ . When $\theta = \frac{\pi}{2}$ , $u = 1$ . The limits of integration change from $\theta = 0$ to $\pi/2$ to $u = 2$ to $\theta$ $1$ .
Evaluate first integral: Change the variable of integration from $\theta$ to $u$ . $-\int_{2}^{1}(\frac{1}{u^2})du$
Substitute $u$ for $\theta$ again: Evaluate the integral of $\frac{1}{u^2}$ with respect to $u$ . $\newline$ $-\int_{2}^{1}\left(\frac{1}{u^2}\right)du = \left[\frac{1}{u}\right]_{2}^{1}$
Simplify integrand: Calculate the value of the integral from the new limits. $\newline$ [\frac{$1$}{u}]_{$2$}^{$1$} = \frac{$1$}{$1$} - \frac{$1$}{$2$} = \frac{$1$}{$2$}\(\newline
Evaluate each integral: For the second integral, use the substitution $u = 1 + \cos(\theta)$ , $du = -\sin(\theta)d\theta$ again. $\newline$ The integral becomes -\int_{$2$}^{$1$}\frac{u^{$2$} - $2$u + $1$}{u^{$2$}} du
Substitute limits: Simplify the integrand and split into separate integrals.\(\newline $-\int_{2}^{1}(1 - 2/u + 1/u^2)\,du$
Combine results: Evaluate each integral separately. $\newline$ $-\int_{2}^{1}du + 2\int_{2}^{1}(\frac{1}{u})du - \int_{2}^{1}(\frac{1}{u^2})du$
Simplify final expression: Calculate the value of each integral. $-[u]_{2}^{1} + 2\cdot[\ln|u|]_{2}^{1} - \left[\frac{1}{u}\right]_{2}^{1}$
Combine like terms: Substitute the limits of integration. $\newline$ $-[(1) - (2)] + 2\cdot[\ln|1| - \ln|2|] - [\left(\frac{1}{1}\right) - \left(\frac{1}{2}\right)]$
Combine like terms: Substitute the limits of integration. $\newline$ $-[(1) - (2)] + 2*[\ln|1| - \ln|2|] - [(\frac{1}{1}) - (\frac{1}{2})]$ Simplify the expression. $\newline$ $-(-1) + 2*(0 - \ln(2)) - (1 - \frac{1}{2})$
Combine like terms: Substitute the limits of integration. $\newline$ $-[(1) - (2)] + 2*[\ln|1| - \ln|2|] - [(\frac{1}{1}) - (\frac{1}{2})]$ Simplify the expression. $\newline$ $-(-1) + 2*(0 - \ln(2)) - (1 - \frac{1}{2})$ Combine the results from both parts of the integral. $\newline$ $\frac{1}{2} - (1 + 2*\ln(2) - \frac{1}{2})$
Combine like terms: Substitute the limits of integration. $\newline$ $-[(1) - (2)] + 2*[\ln|1| - \ln|2|] - [(\frac{1}{1}) - (\frac{1}{2})]$ Simplify the expression. $\newline$ $-(-1) + 2*(0 - \ln(2)) - (1 - \frac{1}{2})$ Combine the results from both parts of the integral. $\newline$ $\frac{1}{2} - (1 + 2*\ln(2) - \frac{1}{2})$ Simplify the final expression. $\newline$ $\frac{1}{2} - 1 - 2*\ln(2) + \frac{1}{2}$
Combine like terms: Substitute the limits of integration. $\newline$ $-[(1) - (2)] + 2*[\ln|1| - \ln|2|] - [(\frac{1}{1}) - (\frac{1}{2})]$ Simplify the expression. $\newline$ $-(-1) + 2*(0 - \ln(2)) - (1 - \frac{1}{2})$ Combine the results from both parts of the integral. $\newline$ $\frac{1}{2} - (1 + 2*\ln(2) - \frac{1}{2})$ Simplify the final expression. $\newline$ $\frac{1}{2} - 1 - 2*\ln(2) + \frac{1}{2}$ Combine like terms to get the final answer. $\newline$ $-2*\ln(2)$