Integration by parts is an integration technique used when an integral contains a product of functions. It comes from the product rule for derivatives and is often written as \(\int u\,dv=uv-\int v\,du\). These problems include common types such as polynomial times exponential, polynomial times trig, logarithmic integrals, repeated integration by parts, definite integrals, and circular integration by parts.
Notes
Practice Problems
\(\textbf{1)}\) \(\displaystyle \int xe^{2x} \,dx\)
The answer is \(\displaystyle \frac{xe^{2x}}{2} \, – \, \frac{e^{2x}}{4}+C\)
\(\,\,\,\,\,u=x, \,\,\, dv=e^{2x}\,dx\)
\(\,\,\,\,\,du=dx, \,\,\, v=\frac{1}{2}e^{2x}\)
\(\,\,\,\,\,\int u \,dv= uv\, – \int v \,du\)
\(\,\,\,\,\,\int x e^{2x}\,dx= \left(x\right)\left(\frac{1}{2}e^{2x}\right) – \int \frac{1}{2}e^{2x} \,dx\)
\(\,\,\,\,\,\int x e^{2x}\,dx= \frac{1}{2}xe^{2x} – \frac{1}{4}e^{2x}+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{xe^{2x}}{2} \, – \, \frac{e^{2x}}{4}+C\)
\(\textbf{2)}\) \(\displaystyle \int \theta \sin{5\theta} \,d\theta\)
The answer is \(\displaystyle -\,\frac{\theta \cos{5\theta}}{5} +\frac{\sin{5\theta}}{25} +C\)
\(\,\,\,\,\,u=\theta, \,\,\, dv=\sin{5\theta} \,d\theta\)
\(\,\,\,\,\,du=d\theta, \,\,\, v=-\frac{1}{5}\cos{5\theta}\)
\(\,\,\,\,\,\int u \,dv= uv\, – \int v \,du\)
\(\,\,\,\,\,\int \theta \sin{5\theta} \,d\theta= \left(\theta\right)\left(-\frac{1}{5}\cos{5\theta}\right) – \int -\frac{1}{5}\cos{5\theta} \,d\theta\)
\(\,\,\,\,\,\int \theta \sin{5\theta} \,d\theta= -\,\frac{\theta \cos{5\theta}}{5} +\frac{1}{5} \int \cos{5\theta} \,d\theta\)
\(\,\,\,\,\,\int \theta \sin{5\theta} \,d\theta= -\,\frac{\theta \cos{5\theta}}{5} +\frac{1}{25} \sin{5\theta}+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle -\,\frac{\theta \cos{5\theta}}{5} +\frac{\sin{5\theta}}{25} +C\)
\(\textbf{3)}\) \(\displaystyle \int x \ln{x} \,dx\)
The answer is \(\displaystyle \frac{x^2\ln{x}}{2}-\,\frac{x^2}{4}+C\)
\(\,\,\,\,\,u=\ln{x}, \,\,\, dv= x \,dx\)
\(\,\,\,\,\,du=\frac{1}{x}\,dx, \,\,\, v=\frac{x^2}{2}\)
\(\,\,\,\,\,\int u \,dv= uv\, – \int v \,du\)
\(\,\,\,\,\,\int x \ln{x} \,dx= \frac{x^2}{2} \ln{x} – \int \frac{x^2}{2}\cdot\frac{1}{x}\,dx\)
\(\,\,\,\,\,\int x \ln{x} \,dx= \frac{x^2}{2} \ln{x} – \frac{1}{2} \int x \,dx\)
\(\,\,\,\,\,\int x \ln{x} \,dx= \frac{x^2}{2} \ln{x} – \frac{x^2}{4} + C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{x^2\ln{x}}{2}-\,\frac{x^2}{4}+C\)
\(\textbf{4)}\) \(\displaystyle \int \frac{\ln{x}}{x^6} \,dx\)
The answer is \(-\displaystyle\frac{\ln{x}}{5x^5} – \frac{1}{25x^5} + C\)
\(\,\,\,\,\,u=\ln{x}, \,\,\, dv=\frac{1}{x^6} \,dx\)
\(\,\,\,\,\,du=\frac{1}{x}\,dx, \,\,\, v=\frac{x^{-5}}{-5}=-\frac{1}{5x^5}\)
\(\,\,\,\,\,\int u \,dv= uv\, – \int v \,du\)
\(\,\,\,\,\,\int \frac{\ln{x}}{x^6} \,dx= \left( \ln{x} \right) \left( -\frac{1}{5x^5} \right) – \int \left( -\frac{1}{5x^5} \right) \left( \frac{1}{x} \right) \,dx\)
\(\,\,\,\,\,\int \frac{\ln{x}}{x^6} \,dx= -\frac{\ln{x}}{5x^5} + \frac{1}{5} \int x^{-6} \,dx\)
\(\,\,\,\,\,\int \frac{\ln{x}}{x^6} \,dx= -\frac{\ln{x}}{5x^5} + \frac{1}{5}\cdot \frac{x^{-5}}{-5} + C\)
\(\,\,\,\,\,\text{The answer is } -\displaystyle\frac{\ln{x}}{5x^5} – \frac{1}{25x^5} + C\)
\(\textbf{5)}\) \(\displaystyle \int e^x \cos(x) \,dx\)
The answer is \(\displaystyle \frac{e^x\sin{x}}{2}+\frac{e^x\cos{x}}{2}+C\)
\(\,\,\,\,\,I=\int e^x\cos{x}\,dx\)
\(\,\,\,\,\,u=e^x, \,\,\, dv=\cos{x}\,dx\)
\(\,\,\,\,\,du=e^x\,dx, \,\,\, v=\sin{x}\)
\(\,\,\,\,\,I=e^x\sin{x}-\int e^x\sin{x}\,dx\)
\(\,\,\,\,\,J=\int e^x\sin{x}\,dx\)
\(\,\,\,\,\,u=e^x, \,\,\, dv=\sin{x}\,dx\)
\(\,\,\,\,\,du=e^x\,dx, \,\,\, v=-\cos{x}\)
\(\,\,\,\,\,J=-e^x\cos{x}+\int e^x\cos{x}\,dx\)
\(\,\,\,\,\,J=-e^x\cos{x}+I\)
\(\,\,\,\,\,I=e^x\sin{x}-\left(-e^x\cos{x}+I\right)\)
\(\,\,\,\,\,I=e^x\sin{x}+e^x\cos{x}-I\)
\(\,\,\,\,\,2I=e^x\sin{x}+e^x\cos{x}\)
\(\,\,\,\,\,I=\frac{e^x\sin{x}}{2}+\frac{e^x\cos{x}}{2}+C\)
\(\textbf{6)}\) \(\displaystyle \int x\cos{x}\,dx\)
The answer is \(x\sin{x}+\cos{x}+C\)
\(\,\,\,\,\,u=x, \,\,\, dv=\cos{x}\,dx\)
\(\,\,\,\,\,du=dx, \,\,\, v=\sin{x}\)
\(\,\,\,\,\,\int u\,dv=uv-\int v\,du\)
\(\,\,\,\,\,\int x\cos{x}\,dx=x\sin{x}-\int \sin{x}\,dx\)
\(\,\,\,\,\,\int x\cos{x}\,dx=x\sin{x}+\cos{x}+C\)
\(\,\,\,\,\,\)The answer is \(x\sin{x}+\cos{x}+C\)
\(\textbf{7)}\) \(\displaystyle \int x\sin{x}\,dx\)
The answer is \(-x\cos{x}+\sin{x}+C\)
\(\,\,\,\,\,u=x, \,\,\, dv=\sin{x}\,dx\)
\(\,\,\,\,\,du=dx, \,\,\, v=-\cos{x}\)
\(\,\,\,\,\,\int u\,dv=uv-\int v\,du\)
\(\,\,\,\,\,\int x\sin{x}\,dx=-x\cos{x}-\int(-\cos{x})\,dx\)
\(\,\,\,\,\,\int x\sin{x}\,dx=-x\cos{x}+\int\cos{x}\,dx\)
\(\,\,\,\,\,\int x\sin{x}\,dx=-x\cos{x}+\sin{x}+C\)
\(\,\,\,\,\,\)The answer is \(-x\cos{x}+\sin{x}+C\)
\(\textbf{8)}\) \(\displaystyle \int x e^x\,dx\)
The answer is \(xe^x-e^x+C\)
\(\,\,\,\,\,u=x, \,\,\, dv=e^x\,dx\)
\(\,\,\,\,\,du=dx, \,\,\, v=e^x\)
\(\,\,\,\,\,\int u\,dv=uv-\int v\,du\)
\(\,\,\,\,\,\int xe^x\,dx=xe^x-\int e^x\,dx\)
\(\,\,\,\,\,\int xe^x\,dx=xe^x-e^x+C\)
\(\,\,\,\,\,\)The answer is \(xe^x-e^x+C\)
\(\textbf{9)}\) \(\displaystyle \int x^2e^x\,dx\)
The answer is \(x^2e^x-2xe^x+2e^x+C\)
\(\,\,\,\,\,u=x^2, \,\,\, dv=e^x\,dx\)
\(\,\,\,\,\,du=2x\,dx, \,\,\, v=e^x\)
\(\,\,\,\,\,\int x^2e^x\,dx=x^2e^x-\int 2xe^x\,dx\)
\(\,\,\,\,\,\int xe^x\,dx=xe^x-e^x\)
\(\,\,\,\,\,\int x^2e^x\,dx=x^2e^x-2\left(xe^x-e^x\right)+C\)
\(\,\,\,\,\,\int x^2e^x\,dx=x^2e^x-2xe^x+2e^x+C\)
\(\,\,\,\,\,\)The answer is \(x^2e^x-2xe^x+2e^x+C\)
\(\textbf{10)}\) \(\displaystyle \int \ln{x}\,dx\)
The answer is \(x\ln{x}-x+C\)
\(\,\,\,\,\,\int \ln{x}\,dx=\int \ln{x}\cdot 1\,dx\)
\(\,\,\,\,\,u=\ln{x}, \,\,\, dv=dx\)
\(\,\,\,\,\,du=\frac{1}{x}\,dx, \,\,\, v=x\)
\(\,\,\,\,\,\int u\,dv=uv-\int v\,du\)
\(\,\,\,\,\,\int \ln{x}\,dx=x\ln{x}-\int x\cdot\frac{1}{x}\,dx\)
\(\,\,\,\,\,\int \ln{x}\,dx=x\ln{x}-\int 1\,dx\)
\(\,\,\,\,\,\int \ln{x}\,dx=x\ln{x}-x+C\)
\(\,\,\,\,\,\)The answer is \(x\ln{x}-x+C\)
\(\textbf{11)}\) \(\displaystyle \int x^2\ln{x}\,dx\)
The answer is \(\displaystyle \frac{x^3\ln{x}}{3}-\frac{x^3}{9}+C\)
\(\,\,\,\,\,u=\ln{x}, \,\,\, dv=x^2\,dx\)
\(\,\,\,\,\,du=\frac{1}{x}\,dx, \,\,\, v=\frac{x^3}{3}\)
\(\,\,\,\,\,\int u\,dv=uv-\int v\,du\)
\(\,\,\,\,\,\int x^2\ln{x}\,dx=\frac{x^3\ln{x}}{3}-\int\frac{x^3}{3}\cdot\frac{1}{x}\,dx\)
\(\,\,\,\,\,\int x^2\ln{x}\,dx=\frac{x^3\ln{x}}{3}-\frac{1}{3}\int x^2\,dx\)
\(\,\,\,\,\,\int x^2\ln{x}\,dx=\frac{x^3\ln{x}}{3}-\frac{x^3}{9}+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{x^3\ln{x}}{3}-\frac{x^3}{9}+C\)
\(\textbf{12)}\) \(\displaystyle \int x^3\ln{x}\,dx\)
The answer is \(\displaystyle \frac{x^4\ln{x}}{4}-\frac{x^4}{16}+C\)
\(\,\,\,\,\,u=\ln{x}, \,\,\, dv=x^3\,dx\)
\(\,\,\,\,\,du=\frac{1}{x}\,dx, \,\,\, v=\frac{x^4}{4}\)
\(\,\,\,\,\,\int x^3\ln{x}\,dx=\frac{x^4\ln{x}}{4}-\int\frac{x^4}{4}\cdot\frac{1}{x}\,dx\)
\(\,\,\,\,\,\int x^3\ln{x}\,dx=\frac{x^4\ln{x}}{4}-\frac{1}{4}\int x^3\,dx\)
\(\,\,\,\,\,\int x^3\ln{x}\,dx=\frac{x^4\ln{x}}{4}-\frac{x^4}{16}+C\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{x^4\ln{x}}{4}-\frac{x^4}{16}+C\)
\(\textbf{13)}\) \(\displaystyle \int x^2\sin{x}\,dx\)
The answer is \(-x^2\cos{x}+2x\sin{x}+2\cos{x}+C\)
\(\,\,\,\,\,u=x^2, \,\,\, dv=\sin{x}\,dx\)
\(\,\,\,\,\,du=2x\,dx, \,\,\, v=-\cos{x}\)
\(\,\,\,\,\,\int x^2\sin{x}\,dx=-x^2\cos{x}+\int 2x\cos{x}\,dx\)
\(\,\,\,\,\,\int x\cos{x}\,dx=x\sin{x}+\cos{x}\)
\(\,\,\,\,\,\int x^2\sin{x}\,dx=-x^2\cos{x}+2\left(x\sin{x}+\cos{x}\right)+C\)
\(\,\,\,\,\,\int x^2\sin{x}\,dx=-x^2\cos{x}+2x\sin{x}+2\cos{x}+C\)
\(\,\,\,\,\,\)The answer is \(-x^2\cos{x}+2x\sin{x}+2\cos{x}+C\)
\(\textbf{14)}\) \(\displaystyle \int x^2\cos{x}\,dx\)
The answer is \(x^2\sin{x}+2x\cos{x}-2\sin{x}+C\)
\(\,\,\,\,\,u=x^2, \,\,\, dv=\cos{x}\,dx\)
\(\,\,\,\,\,du=2x\,dx, \,\,\, v=\sin{x}\)
\(\,\,\,\,\,\int x^2\cos{x}\,dx=x^2\sin{x}-\int2x\sin{x}\,dx\)
\(\,\,\,\,\,\int x\sin{x}\,dx=-x\cos{x}+\sin{x}\)
\(\,\,\,\,\,\int x^2\cos{x}\,dx=x^2\sin{x}-2\left(-x\cos{x}+\sin{x}\right)+C\)
\(\,\,\,\,\,\int x^2\cos{x}\,dx=x^2\sin{x}+2x\cos{x}-2\sin{x}+C\)
\(\,\,\,\,\,\)The answer is \(x^2\sin{x}+2x\cos{x}-2\sin{x}+C\)
\(\textbf{15)}\) \(\displaystyle \int \arctan{x}\,dx\)
The answer is \(x\arctan{x}-\frac{1}{2}\ln\left(x^2+1\right)+C\)
\(\,\,\,\,\,\int \arctan{x}\,dx=\int \arctan{x}\cdot1\,dx\)
\(\,\,\,\,\,u=\arctan{x}, \,\,\, dv=dx\)
\(\,\,\,\,\,du=\frac{1}{1+x^2}\,dx, \,\,\, v=x\)
\(\,\,\,\,\,\int \arctan{x}\,dx=x\arctan{x}-\int\frac{x}{1+x^2}\,dx\)
\(\,\,\,\,\,\int\frac{x}{1+x^2}\,dx=\frac{1}{2}\ln\left(x^2+1\right)\)
\(\,\,\,\,\,\int \arctan{x}\,dx=x\arctan{x}-\frac{1}{2}\ln\left(x^2+1\right)+C\)
\(\,\,\,\,\,\)The answer is \(x\arctan{x}-\frac{1}{2}\ln\left(x^2+1\right)+C\)
\(\textbf{16)}\) \(\displaystyle \int \arcsin{x}\,dx\)
The answer is \(x\arcsin{x}+\sqrt{1-x^2}+C\)
\(\,\,\,\,\,\int \arcsin{x}\,dx=\int \arcsin{x}\cdot1\,dx\)
\(\,\,\,\,\,u=\arcsin{x}, \,\,\, dv=dx\)
\(\,\,\,\,\,du=\frac{1}{\sqrt{1-x^2}}\,dx, \,\,\, v=x\)
\(\,\,\,\,\,\int \arcsin{x}\,dx=x\arcsin{x}-\int\frac{x}{\sqrt{1-x^2}}\,dx\)
\(\,\,\,\,\,\int\frac{x}{\sqrt{1-x^2}}\,dx=-\sqrt{1-x^2}\)
\(\,\,\,\,\,\int \arcsin{x}\,dx=x\arcsin{x}+\sqrt{1-x^2}+C\)
\(\,\,\,\,\,\)The answer is \(x\arcsin{x}+\sqrt{1-x^2}+C\)
\(\textbf{17)}\) \(\displaystyle \int_0^1 xe^x\,dx\)
The answer is \(1\)
\(\,\,\,\,\,u=x, \,\,\, dv=e^x\,dx\)
\(\,\,\,\,\,du=dx, \,\,\, v=e^x\)
\(\,\,\,\,\,\int xe^x\,dx=xe^x-e^x\)
\(\,\,\,\,\,\int_0^1 xe^x\,dx=\left[xe^x-e^x\right]_0^1\)
\(\,\,\,\,\,\int_0^1 xe^x\,dx=\left(e-e\right)-\left(0-e^0\right)\)
\(\,\,\,\,\,\int_0^1 xe^x\,dx=0-(-1)\)
\(\,\,\,\,\,\)The answer is \(1\)
\(\textbf{18)}\) \(\displaystyle \int_1^e \ln{x}\,dx\)
The answer is \(1\)
\(\,\,\,\,\,u=\ln{x}, \,\,\, dv=dx\)
\(\,\,\,\,\,du=\frac{1}{x}\,dx, \,\,\, v=x\)
\(\,\,\,\,\,\int \ln{x}\,dx=x\ln{x}-x\)
\(\,\,\,\,\,\int_1^e \ln{x}\,dx=\left[x\ln{x}-x\right]_1^e\)
\(\,\,\,\,\,\int_1^e \ln{x}\,dx=\left(e\cdot1-e\right)-\left(1\cdot0-1\right)\)
\(\,\,\,\,\,\int_1^e \ln{x}\,dx=0-(-1)\)
\(\,\,\,\,\,\)The answer is \(1\)
\(\textbf{19)}\) \(\displaystyle \int_0^{\pi/2} x\cos{x}\,dx\)
The answer is \(\frac{\pi}{2}-1\)
\(\,\,\,\,\,u=x, \,\,\, dv=\cos{x}\,dx\)
\(\,\,\,\,\,du=dx, \,\,\, v=\sin{x}\)
\(\,\,\,\,\,\int x\cos{x}\,dx=x\sin{x}+\cos{x}\)
\(\,\,\,\,\,\int_0^{\pi/2} x\cos{x}\,dx=\left[x\sin{x}+\cos{x}\right]_0^{\pi/2}\)
\(\,\,\,\,\,\int_0^{\pi/2} x\cos{x}\,dx=\left(\frac{\pi}{2}\cdot1+0\right)-\left(0\cdot0+1\right)\)
\(\,\,\,\,\,\)The answer is \(\frac{\pi}{2}-1\)
\(\textbf{20)}\) \(\displaystyle \int_0^{\pi/2} x\sin{x}\,dx\)
The answer is \(1\)
\(\,\,\,\,\,u=x, \,\,\, dv=\sin{x}\,dx\)
\(\,\,\,\,\,du=dx, \,\,\, v=-\cos{x}\)
\(\,\,\,\,\,\int x\sin{x}\,dx=-x\cos{x}+\sin{x}\)
\(\,\,\,\,\,\int_0^{\pi/2} x\sin{x}\,dx=\left[-x\cos{x}+\sin{x}\right]_0^{\pi/2}\)
\(\,\,\,\,\,\int_0^{\pi/2} x\sin{x}\,dx=\left(-\frac{\pi}{2}\cdot0+1\right)-\left(0+0\right)\)
\(\,\,\,\,\,\)The answer is \(1\)
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