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integration by parts

Integration By Parts

The integration by parts formula is an immediate consequence of the product rule for differentiation. $$(uv)' = u'v + uv' \Rightarrow uv' = (uv)' - u'v$$ Integrating the equation on the right hand side of this statement yields $$ \int uv' =\int (uv)' -\int u'v $$ as the differentiation and integration of the first term on the right hand side cancel each other out, we get $$ \int uv' = uv -\int u'v $$

Integration by parts 1

Evaluate the integral below $$ \int x\sin(2x)dx $$
solution - press button to display
Let $u =x$, let $v' = \sin(x)$, then $u' =1$, $v = -\frac{1}{2}\cos(2x)$ The by parts formula states that $$ \int uv' = uv - \int u'v $$ Substitution of the terms above gives $$ \int x\sin(2x) dx = -\frac{1}{2}x\cos(2x) - \int -\frac{1}{2}\cos(2x)dx $$ Evaluation of the integral on the right hand side yields a final answer of $$ \int x\sin(2x) dx = -\frac{1}{2}x\cos(2x) + \frac{1}{4}\sin(2x) + c $$

Integration by parts 2

By applying the technique of integration by parts twice, evaluate the integral $$ \int x^2\sin(3x) dx $$
solution - press button to display
Let $u = x^2$ then $\frac{du}{dx} = 2x$. Let $\frac{dv}{dx} = \sin(3x)$ then $v = -\frac{1}{3}\cos(3x)$ $$\int x^2\sin(3x)dx = \left(x^2\right)\left(-\frac{1}{3}\cos(3x)\right) - \int \left( 2x\right)\left(-\frac{1}{3}\cos(3x) \right)dx$$

We now need to perform integration by parts again, with $u = 2x,\;\frac{dv}{dx} = -\frac{1}{3}\cos(3x)$

$$ \begin{align} \int -\frac{2x}{3}\cos(3x)dx &= -2\left(\frac{x}{3}\sin(3x) - \int\frac{1}{3}\sin(3x)dx \right) \\ &= -2\left(\frac{x}{3}\sin(3x) + \frac{1}{9}\cos(3x)\right) \end{align} $$

Substitution of this result gives $$ \int x^2\sin(3x)dx = \left(x^2\right)\left(-\frac{1}{3}\cos(3x)\right) + 2\left(\frac{x}{3}\sin(3x) + \frac{1}{9}\cos(3x)\right) + c $$

integration by parts 3

Evaluate the integral $$ \int \ln(x+1)(x+1)^2 dx $$
solution - press button to display
Let $u = \ln(x+1)$, then $\frac{du}{dx} = \frac{1}{x+1}$. Let $\frac{dv}{dx} = (x+1)^2$, then $v = \frac{1}{3}(x+1)^3$ $$ \begin{align} \int \ln(x+1)(x+1)^2 dx &= \frac{1}{3}(x+1)^3\ln(x+1) - \int\frac{1}{x+1}(x+1)^2dx \\ &= \frac{1}{3}(x+1)^3\ln(x+1) - \int (x+1) dx \\ &= \frac{1}{3}(x+1)^3\ln(x+1) - \frac{1}{2}(x+1)^2 +c \end{align} $$