a) int_(0)^((pi)/(2))x sin xdx

Identify Integral: Identify the integral to be solved. We need to evaluate the integral of x sin x with respect to x from 0 to π/2.Use Integration by Parts: Use integration by parts. Integration by parts formula is ∫u dv = uv - ∫v du. Let u = x, which means du = dx. Let dv = sin x dx, which means v = -cos x.Apply Integration by Parts: Apply the integration by parts formula. ∫x sin x dx = -x cos x - ∫(-cos x) dxIntegrate -cos x: Integrate -cos x with respect to x. ∫(-cos x) dx = -sin xSubstitute Integrated Parts: Substitute the integrated parts into the formula. ∫x sin x dx = -x cos x + sin x + C, where C is the constant of integration.Evaluate Definite Integral: Evaluate the definite integral from 0 to π/2. ∫_(0)^(π/2) x sin x dx = [-x cos x + sin x] from 0 to π/2 = [-(π/2) cos(π/2) + sin(π/2)] - [-(0) cos(0) + sin(0)] = [-(π/2)(0) + 1] - [0 + 0] = 1

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a) $\int_{0}^{\frac{\pi}{2}} x \sin x \, dx$

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Calculus

Evaluate definite integrals using the chain rule

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Q. a)

\int_{0}^{\frac{\pi}{2}} x \sin x \, dx

Identify Integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of $x \sin x$ with respect to $x$ from $0$ to $\frac{\pi}{2}$ .
Use Integration by Parts: Use integration by parts. $\newline$ Integration by parts formula is $\int u \, dv = uv - \int v \, du$ . $\newline$ Let $u = x$ , which means $du = dx$ . $\newline$ Let $dv = \sin x \, dx$ , which means $v = -\cos x$ .
Apply Integration by Parts: Apply the integration by parts formula. $\int x \sin x \, dx = -x \cos x - \int(-\cos x) \, dx$
Integrate $-\cos x$ : Integrate $-\cos x$ with respect to $x$ . $\int(-\cos x) \, dx = -\sin x$
Substitute Integrated Parts: Substitute the integrated parts into the formula. $\newline$ $\int x \sin x \, dx = -x \cos x + \sin x + C$ , where $C$ is the constant of integration.
Evaluate Definite Integral: Evaluate the definite integral from $0$ to $\frac{\pi}{2}$ .
$\int_{0}^{\frac{\pi}{2}} x \sin x \,dx = [-x \cos x + \sin x]$ from $0$ to $\frac{\pi}{2}$
= $[-(\frac{\pi}{2}) \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})] - [-(0) \cos(0) + \sin(0)]$
= $[-(\frac{\pi}{2})(0) + 1] - [0 + 0]$
= $1$