Calculus BC ยท Unit 6
Choose u so the new integral is simpler, then evaluate the boundary term on a definite integral.
The parts formula
\(\int u\,dv = uv-\int v\,du\)
Differentiate \(u\). Integrate \(dv\).
Choose u so that du gets simpler
A useful order is logarithm, inverse trig, algebraic, trig, exponential.
Put the earlier type in \(u\).
A definite integral uses the same formula
Evaluate the \(uv\) boundary term from the lower limit to the upper limit, then evaluate the new integral.
Sometimes parts is used twice
A polynomial times \(e^x\) or \(\sin x\) may need one parts step for each degree.
One parts step. The new integral is basic.
\(u=x,\quad dv=e^x\,dx\)
\(du=dx,\quad v=e^x\)
\(uv-\int v\,du\)
\(xe^x-\int e^x\,dx\)
\(xe^x-e^x+C\)
\(e^x(x-1)+C\)
Answer: \(e^x(x-1)+C\)
\(u=\ln x,\quad dv=dx\)
\(du=\dfrac{1}{x}\,dx,\quad v=x\)
\(x\ln x-\int 1\,dx\)
\(x\ln x-x+C\)
Answer: \(x\ln x-x+C\)
An antiderivative is \(e^x(x-1)\).
\(\left[e^x(x-1)\right]_0^1\)
At 1: \(e\cdot 0=0\).
At 0: \(1\cdot(-1)=-1\).
\(0-(-1)\)
\(1\)
Answer: \(1\)
\(u=x,\quad dv=\cos x\,dx\)
\(du=dx,\quad v=\sin x\)
\(x\sin x-\int \sin x\,dx\)
\(x\sin x+\cos x+C\)
Answer: \(x\sin x+\cos x+C\)
The algebraic factor should be \(u\), because its derivative is a constant.
Integrating \(e^{2x}\) stays in the same family.
Choosing \(u=e^{2x}\) raises the power of \(x\) and makes the work longer.
Answer: A. \(u=x\) and \(dv=e^{2x}\,dx\)
An antiderivative is \(x\ln x-x\).
\(\left[x\ln x-x\right]_1^e\)
At \(e\): \(e\cdot 1-e=0\).
At 1: \(0-1=-1\).
\(0-(-1)\)
\(1\)
Answer: \(1\)
A coefficient appears when v is an exponential, or the integrand is x sin x.
\(u=x,\quad dv=\sin x\,dx\)
\(du=dx,\quad v=-\cos x\)
\(-x\cos x-\int -\cos x\,dx\)
\(-x\cos x+\int \cos x\,dx\)
\(-x\cos x+\sin x+C\)
Answer: \(-x\cos x+\sin x+C\)
\(u=x,\quad dv=e^{2x}\,dx\)
\(du=dx,\quad v=\dfrac{1}{2}e^{2x}\)
\(\dfrac{x}{2}e^{2x}-\int \dfrac{1}{2}e^{2x}\,dx\)
\(\dfrac{x}{2}e^{2x}-\dfrac{1}{4}e^{2x}+C\)
\(\dfrac{e^{2x}}{4}(2x-1)+C\)
Answer: \(\dfrac{e^{2x}}{4}(2x-1)+C\)
An antiderivative is \(-x\cos x+\sin x\).
\(\left[-x\cos x+\sin x\right]_0^{\pi}\)
At \(\pi\): \(-\pi\cos\pi+\sin\pi\).
\(-\pi(-1)\)
\(\pi\)
At 0: \(-0\cdot 1+0=0\).
\(\pi-0\)
\(\pi\)
Answer: \(\pi\)
\(u=\arctan x,\quad dv=dx\)
\(du=\dfrac{1}{1+x^2}\,dx,\quad v=x\)
\(x\arctan x-\int \dfrac{x}{1+x^2}\,dx\)
For the new integral, \(w=1+x^2\) and \(dw=2x\,dx\).
\(\dfrac{1}{2}\ln(1+x^2)\)
\(x\arctan x-\dfrac{1}{2}\ln(1+x^2)+C\)
Answer: \(x\arctan x-\dfrac{1}{2}\ln(1+x^2)+C\)
An antiderivative is \(\dfrac{e^{2x}}{4}(2x-1)\).
At 1: \(\dfrac{e^2}{4}(1)\).
At 0: \(\dfrac{1}{4}(-1)=-\dfrac{1}{4}\).
\(\dfrac{e^2}{4}-\left(-\dfrac{1}{4}\right)\)
\(\dfrac{e^2+1}{4}\)
Answer: \(\dfrac{e^2+1}{4}\)
\(u=\ln x,\quad dv=x\,dx\)
\(du=\dfrac{1}{x}\,dx,\quad v=\dfrac{x^2}{2}\)
\(\dfrac{x^2}{2}\ln x-\int \dfrac{x}{2}\,dx\)
\(\dfrac{x^2}{2}\ln x-\dfrac{x^2}{4}+C\)
Answer: \(\dfrac{x^2}{2}\ln x-\dfrac{x^2}{4}+C\)
Two rounds of parts, or a logarithm times a power.
\(u=x^2,\quad dv=e^x\,dx\)
\(x^2e^x-\int 2x e^x\,dx\)
Use parts again on \(\int x e^x\,dx=e^x(x-1)\).
\(x^2e^x-2e^x(x-1)+C\)
\(e^x(x^2-2x+2)+C\)
Answer: \(e^x(x^2-2x+2)+C\)
An antiderivative is \(e^x(x^2-2x+2)\).
At 1: \(e(1-2+2)=e\).
At 0: \(1\cdot 2=2\).
\(e-2\)
Answer: \(e-2\)
\(u=\ln x,\quad dv=x^2\,dx\)
\(du=\dfrac{1}{x}\,dx,\quad v=\dfrac{x^3}{3}\)
\(\dfrac{x^3}{3}\ln x-\int \dfrac{x^2}{3}\,dx\)
\(\dfrac{x^3}{3}\ln x-\dfrac{x^3}{9}+C\)
Answer: \(\dfrac{x^3}{3}\ln x-\dfrac{x^3}{9}+C\)
An antiderivative is \(\dfrac{x^3}{3}\ln x-\dfrac{x^3}{9}\).
At \(e\): \(\dfrac{e^3}{3}-\dfrac{e^3}{9}\).
\(\dfrac{3e^3-e^3}{9}\)
\(\dfrac{2e^3}{9}\)
At 1: \(0-\dfrac{1}{9}=-\dfrac{1}{9}\).
\(\dfrac{2e^3}{9}-\left(-\dfrac{1}{9}\right)\)
\(\dfrac{2e^3+1}{9}\)
Answer: \(\dfrac{2e^3+1}{9}\)
An antiderivative is \(x\sin x+\cos x\).
At \(\pi/2\): \(\dfrac{\pi}{2}\cdot 1+0\).
\(\dfrac{\pi}{2}\)
At 0: \(0+1=1\).
\(\dfrac{\pi}{2}-1\)
Answer: \(\dfrac{\pi}{2}-1\)
The factor \(x\) is not the derivative of the inside of \(\cos x\).
No choice of \(u\) turns the whole product into a basic \(\int f(u)\,du\).
Parts with \(u=x\) and \(dv=\cos x\,dx\) leaves \(\int \sin x\,dx\).
That new integral is basic.
Answer: use parts with \(u=x\)
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