Integration by Substitution

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\(\int f(u(x))\,u'(x)\,dx = \int f(u)\,du\)

What to Remember

The extra factor should be du

Choose \(u\) as the inside function. Compute \(du\). The remaining factor in the integral should match \(du\), up to a constant.

Change the limits on a definite integral

When \(x=a\), use \(u(a)\). When \(x=b\), use \(u(b)\). Do not switch back to \(x\) at the end if the limits are already \(u\)-values.

A lone constant moves outside

If \(du=6x\,dx\) but the integral has \(x\,dx\), write \(x\,dx=\dfrac{1}{6}\,du\).

Look for a derivative hiding in the numerator

\(\int \dfrac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C\) when that pattern is really there.

Level 1

Easy

The derivative of the inside is already in the integral.

Example

Find \(\int 2x(x^2+3)^4\,dx\).

\(u=x^2+3\)

\(du=2x\,dx\)

\(\int u^4\,du\)

\(\dfrac{u^5}{5}+C\)

\(\dfrac{(x^2+3)^5}{5}+C\)

Answer: \(\dfrac{(x^2+3)^5}{5}+C\)

Find \(\int 3\cos(3x)\,dx\).

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\(u=3x\)

\(du=3\,dx\)

\(\int \cos u\,du\)

\(\sin(3x)+C\)

Answer: \(\sin(3x)+C\)

Find \(\int e^{4x}\,dx\).

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\(u=4x\)

\(du=4\,dx\)

\(dx=\dfrac{1}{4}\,du\)

\(\dfrac{1}{4}e^{4x}+C\)

Answer: \(\dfrac{1}{4}e^{4x}+C\)

Find \(\int \dfrac{2x}{x^2+5}\,dx\).

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\(u=x^2+5\)

\(du=2x\,dx\)

\(\int \dfrac{1}{u}\,du\)

\(\ln|x^2+5|+C\)

Answer: \(\ln(x^2+5)+C\)

Find \(\int_0^2 2x(x^2+1)^2\,dx\).

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\(u=x^2+1\)

\(du=2x\,dx\)

When \(x=0\), \(u=1\). When \(x=2\), \(u=5\).

\(\int_1^5 u^2\,du\)

\(\left[\dfrac{u^3}{3}\right]_1^5\)

\(\dfrac{125}{3}-\dfrac{1}{3}\)

\(\dfrac{124}{3}\)

Answer: \(\dfrac{124}{3}\)

Find \(\int \sin x\cos x\,dx\).

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\(u=\sin x\)

\(du=\cos x\,dx\)

\(\int u\,du\)

\(\dfrac{\sin^2 x}{2}+C\)

Answer: \(\dfrac{\sin^2 x}{2}+C\)

Level 2

Medium

A missing constant, a root, or a definite integral.

Example

Find \(\int x\sqrt{x^2+4}\,dx\).

\(u=x^2+4\)

\(du=2x\,dx\)

\(x\,dx=\dfrac{1}{2}\,du\)

\(\dfrac{1}{2}\int u^{1/2}\,du\)

\(\dfrac{1}{2}\cdot \dfrac{u^{3/2}}{3/2}\)

\(\dfrac{1}{2}\cdot \dfrac{2}{3}u^{3/2}\)

\(\dfrac{1}{3}(x^2+4)^{3/2}+C\)

Answer: \(\dfrac{1}{3}(x^2+4)^{3/2}+C\)

Find \(\int_0^1 3x^2(x^3+1)^2\,dx\).

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\(u=x^3+1\)

\(du=3x^2\,dx\)

Limits: \(u(0)=1\) and \(u(1)=2\).

\(\int_1^2 u^2\,du\)

\(\left[\dfrac{u^3}{3}\right]_1^2\)

\(\dfrac{8}{3}-\dfrac{1}{3}\)

\(\dfrac{7}{3}\)

Answer: \(\dfrac{7}{3}\)

Find \(\int \dfrac{\ln x}{x}\,dx\) for \(x>1\).

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\(u=\ln x\)

\(du=\dfrac{1}{x}\,dx\)

\(\int u\,du\)

\(\dfrac{(\ln x)^2}{2}+C\)

Answer: \(\dfrac{(\ln x)^2}{2}+C\)

Find \(\int_0^{\pi/6} \sin(2x)\,dx\).

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\(u=2x\)

\(du=2\,dx\)

\(dx=\dfrac{1}{2}\,du\)

When \(x=0\), \(u=0\). When \(x=\pi/6\), \(u=\pi/3\).

\(\dfrac{1}{2}\int_0^{\pi/3}\sin u\,du\)

\(\dfrac{1}{2}\left[-\cos u\right]_0^{\pi/3}\)

\(\dfrac{1}{2}\left(-\dfrac{1}{2}-(-1)\right)\)

\(\dfrac{1}{2}\cdot \dfrac{1}{2}\)

\(\dfrac{1}{4}\)

Answer: \(\dfrac{1}{4}\)

Find \(\int \tan x\,dx\) on an interval where \(\cos x>0\).

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\(\tan x=\dfrac{\sin x}{\cos x}\)

\(u=\cos x\)

\(du=-\sin x\,dx\)

\(-\int \dfrac{1}{u}\,du\)

\(-\ln|\cos x|+C\)

Answer: \(-\ln(\cos x)+C\)

Find \(\int_0^1 \dfrac{x}{x^2+1}\,dx\).

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\(u=x^2+1\)

\(du=2x\,dx\)

Limits: \(u(0)=1\) and \(u(1)=2\).

\(\dfrac{1}{2}\int_1^2 \dfrac{1}{u}\,du\)

\(\dfrac{1}{2}\left[\ln u\right]_1^2\)

\(\dfrac{1}{2}(\ln 2-\ln 1)\)

\(\dfrac{1}{2}\ln 2\)

Answer: \(\dfrac{1}{2}\ln 2\)

Level 3

Hard

A definite integral whose limits change, or a power that is not the inside.

Example

Find \(\int x^2(x^3-1)^5\,dx\).

\(u=x^3-1\)

\(du=3x^2\,dx\)

\(x^2\,dx=\dfrac{1}{3}\,du\)

\(\dfrac{1}{3}\int u^5\,du\)

\(\dfrac{1}{3}\cdot \dfrac{u^6}{6}+C\)

\(\dfrac{(x^3-1)^6}{18}+C\)

Answer: \(\dfrac{(x^3-1)^6}{18}+C\)

Find \(\int_1^e \dfrac{\ln x}{x}\,dx\).

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\(u=\ln x\)

Limits: \(u(1)=0\) and \(u(e)=1\).

\(\int_0^1 u\,du\)

\(\left[\dfrac{u^2}{2}\right]_0^1\)

\(\dfrac{1}{2}\)

Answer: \(\dfrac{1}{2}\)

Find \(\int \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\,dx\) for \(x>0\).

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\(u=\sqrt{x}\)

\(du=\dfrac{1}{2\sqrt{x}}\,dx\)

\(\dfrac{1}{\sqrt{x}}\,dx = 2\,du\)

\(2\int e^u\,du\)

\(2e^{\sqrt{x}}+C\)

Answer: \(2e^{\sqrt{x}}+C\)

Find \(\int_0^{\sqrt{3}} \dfrac{1}{1+x^2}\,dx\) without a substitution. Then explain why a substitution is the wrong first move.

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This matches \(\int \dfrac{1}{1+x^2}\,dx=\arctan x\) directly.

\(\left[\arctan x\right]_0^{\sqrt{3}}\)

\(\arctan\sqrt{3}-\arctan 0\)

\(\dfrac{\pi}{3}-0\)

\(\dfrac{\pi}{3}\)

There is no extra factor that looks like the derivative of an inside function.

Answer: \(\dfrac{\pi}{3}\)

Evaluate \(\int_1^2 \dfrac{2x+2}{x^2+2x}\,dx\).

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The numerator is the derivative of the denominator.

\(u=x^2+2x\)

\(du=(2x+2)\,dx\)

When \(x=1\), \(u=3\). When \(x=2\), \(u=8\).

\(\int_3^8 \dfrac{1}{u}\,du\)

\(\ln 8-\ln 3\)

\(\ln\dfrac{8}{3}\)

Answer: \(\ln\dfrac{8}{3}\)

A student evaluates \(\int_0^1 2x(x^2+1)^2\,dx\) and writes \(\left[\dfrac{(x^2+1)^3}{3}\right]_0^1\) but forgets to use the new limits, getting \(\dfrac{8}{3}-\dfrac{1}{3}=\dfrac{7}{3}\). That number happens to be right. Explain the two correct paths.

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Path 1: change limits. \(u=x^2+1\) runs from 1 to 2.

\(\int_1^2 u^2\,du=\dfrac{8}{3}-\dfrac{1}{3}=\dfrac{7}{3}\)

Path 2: return to x before substituting the original limits.

\(\left[\dfrac{(x^2+1)^3}{3}\right]_0^1=\dfrac{8}{3}-\dfrac{1}{3}=\dfrac{7}{3}\)

Both paths agree. Mixing new antiderivatives with old limits is the step that usually fails.

Answer: \(\dfrac{7}{3}\)

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