Tag: AP Calculus

Integration by parts serves as a method to streamline integrals structured as $latex \int f\left(x\right)g\left(x\right) dx$. The method becomes beneficial when function $latex f$ can undergo repeated differentiation and function $latex g$ can be integrated repeatedly without encountering challenges. For example, $latex \displaystyle\int x\sin x dx$ $latex \displaystyle\int xe^xdx$ The formula: $latex \displaystyle\int f\left(x\right)g'\left(x\right)dx=f\left(x\right)g\left(x\right)-\int f'\left(x\right)g\left(x\right)dx$ or $latex \displaystyle\int udv=uv-\int vdu$ Integration by Parts Proof: The Product Rule says that if $latex f\left(x\right)$ and $latex g\left(x\right)$ are both differentiable functions, then $latex \displaystyle\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]=f\left(x\right)g'\left(x\right)$ Rewriting this using the notation for indefinite integrals: $latex \displaystyle\int \left[f\left(x\right)g'\left(x\right)+g\left(x\right)f'\left(x\right)\right]dx=f\left(x\right)g\left(x\right)$ or $latex \displaystyle\int f\left(x\right)g'\left(x\right)dx+\int g\left(x\right)f'\left(x\right)dx=f\left(x\right)g\left(x\right)$ Arranging the terms we obtain $latex \displaystyle\int f\left(x\right)g'\left(x\right)dx=f\left(x\right)g\left(x\right)-\int g\left(x\right)f'\left(x\right)$ Let $latex u=f\left(x\right)$ and $latex v=g\left(x\right)$: $latex du=f'\left(x\right)dx$ and $latex dv=g'\left(x\right)dx$ Then, by substitution, $latex \displaystyle \int udv=uv-\int vdu$ Example 1 Find $latex \displaystyle\int x\sin xdx$. Solution Choose $latex f\left(x\right)=x$ and $latex g'\left(x\right)=\sin x$: $latex \begin{array}{rcl}\displaystyle\int x\sin xdx&=&f\left(x\right)g\left(x\right)-\int g\left(x\right)f'\left(x\right)dx \\ &=& x\left(-\cos x\right)-\displaystyle\int\left(-\cos x\right)dx \\ &=& -x\cos x+\displaystyle\int\cos xdx \\ &=& -x\cos x+\sin x+C \end{array}$ Example 2 Evaluate $latex \displaystyle\int\ln xdx$. Solution Let $latex u=\ln x$ and $latex dv=dx$: $latex \displaystyle u=\ln x\rightarrow du=\frac{1}{x}dx$ $latex \displaystyle dv=dx\rightarrow v=x$ Now, let's integrate by parts: $latex \begin{array}{rcl} \displaystyle\int\ln xdx &=& x\ln x-\int x\cdot\frac{1}{x}dx \\ &=& x\ln x-\displaystyle\int dx \\ &=& x\ln x-x+C \end{array}$ Example 3 Find $latex \displaystyle t^2e^tdt$. Solution Let $latex u=t^2$ and $latex dv=e^tdt$: $latex \begin{array}{rcl} \displaystyle\int t^2e^tdt &=& t^2e^t-2\int te^tdt \\ &=& t^2e^t-2\left(te^t-\displaystyle\int e^tdt\right) \\ &=& t^2e^t-2\left(te^t-e^t+C\right) \\ &=& t^2e^t-2te^t+2e^t+C_1 \end{array}$ where $latex C_1=-2C$. Example 4 Evaluate $latex \displaystyle\int e^x\sin xdx$. Solution Let's first choose $latex u=e^x$ and $latex dv=\sin x dx$: $latex \displaystyle \int e^x\sin xdx=-e^x\cos x+\int e^x\cos xdx$ But $latex \displaystyle\int e^x\cos xdx=e^x\sin x-\int e^x\sin xdx$ Therefore, $latex \displaystyle\int e^x\sin xdx=-e^x\cos x+e^x\sin x-\int e^x\sin xdx$ Let's add $latex \displaystyle\int e^x\sin xdx$ to both sides of the equation: $latex \displaystyle 2\int e^x\sin xdx=-e^x\cos x+e^x\sin x$ $latex \displaystyle\int e^x\sin xdx=\frac{1}{2}e^x\left(\sin x-\cos x\right)+C$ Example 5 Calculate $latex \displaystyle\int_0^{1}\tan^{-1} xdx$. Solution Let $latex u=\tan^{-1}x$ and $latex dv=dx$: $latex \displaystyle du=\frac{dx}{1+x^2}\qquad v=x$ Integrate: $latex \begin{array}{rcl} \int_0^1\tan^{-1}xdx &=& x\tan^{-1}x\Bigr|_0^1-\int_0^1\frac{x}{1+x^2}dx \\ &=& \displaystyle 1\cdot \tan^{-1}1-0\cdot \tan^{-1}0-\int_0^1\frac{x}{1+x^2}dx \\ &=& \displaystyle\frac{\pi}{4}-\int_0^1\frac{x}{1+x^2}dx \\ &=& \displaystyle \frac{\pi}{4}-\frac{\ln 2}{2} \end{array}$ To evaluate $latex \displaystyle \int\frac{x}{1+x^2}dx$, use the substitution $latex k=1+x^2$. More Practice Questions