$\int (x'Co_{3}2+2) |4-x| dx$

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Algebra 2

Transformations of absolute value functions

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\int (x'Co_{3}2+2) |4-x| dx

Identify Integral: Identify the integral to solve. $\int(x^3 \cdot \cos(2x) + 2) \, dx$
Split Integral: Split the integral into two parts. $\int x^3 \cdot \cos(2x) \, dx + \int 2 \, dx$
Integrate Constant: Integrate the constant $2$ with respect to $x$ . $\int 2 \, dx = 2x$
Apply Integration by Parts: Use integration by parts for $\int x^3 \cdot \cos(2x) \, dx$ . Let $u = x^3$ , $dv = \cos(2x) \, dx$ Then $du = 3x^2 \, dx$ , $v = (1/2)\sin(2x)$
Simplify Expression: Apply the integration by parts formula: $\int u\,dv = uv - \int v\,du$ .\int x^\(3 \cos( $2$ x) \,dx = \left(x^ $3$ \left(\frac{ $1$ }{ $2$ }\right)\sin( $2$ x)\right) - \int\left(\left(\frac{ $1$ }{ $2$ }\right)\sin( $2$ x) \cdot $3$ x^ $2$ \,dx\right)
Use Integration by Parts Again: Simplify the expression. $\newline$ $x^3 \cdot (\frac{1}{2})\sin(2x)$ - $(\frac{3}{2})\int x^2 \cdot \sin(2x) \, dx$
Integrate Remaining Part: Use integration by parts again for $\int x^2 \sin(2x) \, dx$ . Let $u = x^2$ , $dv = \sin(2x) \, dx$ Then $du = 2x \, dx$ , $v = -(\frac{1}{2})\cos(2x)$
Integrate Last Part: Apply the integration by parts formula again. $\newline$ $(\frac{3}{2})\int x^2 \sin(2x) \, dx = (\frac{3}{2})(x^2 \cdot -(\frac{1}{2})\cos(2x)) - (\frac{3}{2})\int(-(\frac{1}{2})\cos(2x) \cdot 2x \, dx)$
Combine All Parts: Simplify the expression. $\newline$ $(\frac{3}{2})(x^2 * -(\frac{1}{2})\cos(2x)) - (\frac{3}{4})\int(-\cos(2x) * 2x \, dx)$
Combine All Parts: Simplify the expression. $\newline$ $(\frac{3}{2})(x^2 * -(\frac{1}{2})\cos(2x)) - (\frac{3}{4})\int(-\cos(2x) * 2x \, dx)$ Integrate the remaining part by parts again. $\newline$ Let $u = 2x$ , $dv = -\cos(2x) \, dx$ $\newline$ Then $du = 2 \, dx$ , $v = -(\frac{1}{2})\sin(2x)$
Combine All Parts: Simplify the expression. $\newline$ $(\frac{3}{2})(x^2 * -(\frac{1}{2})\cos(2x)) - (\frac{3}{4})\int(-\cos(2x) * 2x \, dx)$ Integrate the remaining part by parts again. $\newline$ Let $u = 2x$ , $dv = -\cos(2x) \, dx$ $\newline$ Then $du = 2 \, dx$ , $v = -(\frac{1}{2})\sin(2x)$ Apply the integration by parts formula one more time. $\newline$ $(\frac{3}{4})\int(-\cos(2x) * 2x \, dx) = (\frac{3}{4})(2x * -(\frac{1}{2})\sin(2x)) - (\frac{3}{4})\int(-(\frac{1}{2})\sin(2x) * 2 \, dx)$
Combine All Parts: Simplify the expression. $\newline$ $(\frac{3}{2})(x^2 * -(\frac{1}{2})\cos(2x)) - (\frac{3}{4})\int(-\cos(2x) * 2x \, dx)$ Integrate the remaining part by parts again. $\newline$ Let $u = 2x$ , $dv = -\cos(2x) \, dx$ $\newline$ Then $du = 2 \, dx$ , $v = -(\frac{1}{2})\sin(2x)$ Apply the integration by parts formula one more time. $\newline$ $(\frac{3}{4})\int(-\cos(2x) * 2x \, dx) = (\frac{3}{4})(2x * -(\frac{1}{2})\sin(2x)) - (\frac{3}{4})\int(-(\frac{1}{2})\sin(2x) * 2 \, dx)$ Simplify and integrate the last part. $\newline$ $(\frac{3}{4})(2x * -(\frac{1}{2})\sin(2x)) - (\frac{3}{4})(-1)\int\sin(2x) \, dx$
Combine All Parts: Simplify the expression. $\newline$ $(\frac{3}{2})(x^2 * -(\frac{1}{2})\cos(2x)) - (\frac{3}{4})\int(-\cos(2x) * 2x \, dx)$ Integrate the remaining part by parts again. $\newline$ Let $u = 2x$ , $dv = -\cos(2x) \, dx$ $\newline$ Then $du = 2 \, dx$ , $v = -(\frac{1}{2})\sin(2x)$ Apply the integration by parts formula one more time. $\newline$ $(\frac{3}{4})\int(-\cos(2x) * 2x \, dx) = (\frac{3}{4})(2x * -(\frac{1}{2})\sin(2x)) - (\frac{3}{4})\int(-(\frac{1}{2})\sin(2x) * 2 \, dx)$ Simplify and integrate the last part. $\newline$ $(\frac{3}{4})(2x * -(\frac{1}{2})\sin(2x)) - (\frac{3}{4})(-1)\int\sin(2x) \, dx$ Integrate $\int\sin(2x) \, dx$ . $\newline$ $\int\sin(2x) \, dx = -(\frac{1}{2})\cos(2x)$
Combine All Parts: Simplify the expression. $\newline$ $(\frac{3}{2})(x^2 * -(\frac{1}{2})\cos(2x)) - (\frac{3}{4})\int(-\cos(2x) * 2x \, dx)$ Integrate the remaining part by parts again. $\newline$ Let $u = 2x$ , $dv = -\cos(2x) \, dx$ $\newline$ Then $du = 2 \, dx$ , $v = -(\frac{1}{2})\sin(2x)$ Apply the integration by parts formula one more time. $\newline$ $(\frac{3}{4})\int(-\cos(2x) * 2x \, dx) = (\frac{3}{4})(2x * -(\frac{1}{2})\sin(2x)) - (\frac{3}{4})\int(-(\frac{1}{2})\sin(2x) * 2 \, dx)$ Simplify and integrate the last part. $\newline$ $(\frac{3}{4})(2x * -(\frac{1}{2})\sin(2x)) - (\frac{3}{4})(-1)\int\sin(2x) \, dx$ Integrate $\int\sin(2x) \, dx$ . $\newline$ $\int\sin(2x) \, dx = -(\frac{1}{2})\cos(2x)$ Combine all parts together and add the constant of integration $C$ . $\newline$ $u = 2x$ $0$