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Section \(u\) -substitution
When an antiderivative is not immediately obvious, we sometimes use a substitution. Substitution is a good choice when we have one part of the integrand "looks like" the derivative of another part of the integrand.
π Letβs look a a few examples
π Example 323 . Compute \(\ds \int \, \cos x\sin x \ dx\) .
Since the derivative of
\(\cos x\) is
\(-\sin x\) which "looks like"
\(\sin x\text{,}\) then
\(u\) -substitution (with
\(u=\cos x\) ) works well.
Note, \(u=\sin x\) would also work.
\begin{align*}
u \amp = \cos x \\
du \amp = -\sin x \, dx \\
-\frac{1}{\sin x}du \amp = dx
\end{align*}
\begin{align*}
\int \, \cos x\sin x \ dx \amp = \int u \sin x \,
\left[-\frac{1}{\sin x}du\right] \\
\amp = -\int u \,du \\
\amp = -\frac{u^2}{2} + C \\
\amp = -\frac{1}{2}\cos^2 x + C
\end{align*}
π Example 324 . Compute \(\ds \int \frac{1}{x}\cdot \ln(15x) dx\) .
Substitution is a good choice here because we have a natural log function and we have
\(\ds \frac{1}{x},\) which is the derivative of natural log. So we would choose
\begin{align*}
u \amp = \ln(15x) \\
du \amp = \frac{d}{dx}\Big( \ln(15x) \Big) \\
\amp = \frac{1}{15x}\cdot \frac{d}{dx}(15x) \\
\amp = \frac{1}{15x}\cdot 15dx \\
\amp = \frac{1}{x}du
\end{align*}
\begin{align*}
\int \frac{1}{x}\cdot \ln(15x) dx
\amp = \int \ub{\ln(15x)}_{u} \cdot \ub{\frac{1}{x} dx}_{du} \\
\amp = \int udu \\
\amp = \frac{1}{2}u^2 + C \\
\amp = \frac{1}{2}( \ln(15x))^2 + C
\end{align*}
We usually pause to confirm that the new integral is simpler than the original. In this case it certainly is, so we proceed to integrate and then change back to the original variable,
\(x\text{.}\)
Evaluate each of the following integrals. Use proper notation.
\(\ds \int e^z \cos(e^z) dz \qquad\) Solution .
We will choose \(u = e^z.\) Then \(du = e^z dz.\) Therefore we have:
\begin{align*}
\int e^z \cos(e^z) dz
\amp = \int \cos(e^z) e^z dz \\
\amp = \int \ub{\cos(e^z)}_{ss \cos (u)} \ub{e^z dz}_{ss du} \\
\amp = \int \cos(u) du \\
\amp = \sin(u) + C \\
\amp = \sin(e^z) + C
\end{align*}
Answer .
\begin{equation*}
\sin(e^z) + C
\end{equation*}
\(\ds \int t(t^2 - 12)^{57} dt \qquad\) Solution .
We will choose \(u = t^2 - 12.\) Then \(du = 2tdt,\) and therefore \(\ds\frac{1}{2}du = tdt\) Thus we have:
\begin{align*}
\int t(t^2 - 12)^{57} dt
\amp = \int (t^2 - 12)^{57} tdt \\
\amp = \int \ub{(t^2 - 12)^{57}}_{ss u^{57}}
\ub{tdt}_{ss \frac{1}{2}du} \\
\amp = \int u^{57}\cdot \frac{1}{2}du \\
\amp = \frac{1}{2}\int u^{57}du \\
\amp = \frac{1}{2}\cdot \frac{1}{58} u^{58} + C \\
\amp = \frac{1}{116}(t^2 - 12)^{58} + C
\end{align*}
Answer .
\begin{equation*}
\frac{1}{116}(t^2 - 12)^{58} + C
\end{equation*}
\(\ds \int 5x^2 e^{(7x^3)} dx \qquad\) Solution .
We will choose \(u = 7x^3.\) Then \(du = 21x^2dx,\) and therefore \(\ds\frac{1}{21}du = x^2dx\) Thus we have:
\begin{align*}
\int 5x^2 e^{(7x^3)} dx
\amp = 5 \int e^{(7x^3)} x^2 dx \\
\amp = 5\int \ub{e^{(7x^3)}}_{ss e^u}
\ub{x^2 dx}_{ss \frac{1}{21}du} \\
\amp = 5\int e^u\cdot \frac{1}{21}du \\
\amp = \frac{5}{21}\int e^u du \\
\amp = \frac{5}{21}e^u + C \\
\amp = \frac{5}{21}e^{(7x^3)} + C
\end{align*}
Answer .
\begin{equation*}
\frac{5}{21}e^{(7x^3)} + C
\end{equation*}
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