Exercise 10.3
Use the integration by parts to evaluate the following integrals.
(a) $\int s e^{-2s} ds$
Solution:
Let $u = s, \quad dv = e^{-2s} ds$
$du = ds, \quad v = -\frac{1}{2} e^{-2s}$
$$ \begin{aligned}
\int s e^{-2s} ds &= s(-\frac{1}{2} e^{-2s}) - \int (-\frac{1}{2} e^{-2s}) ds \\
&= -\frac{1}{2} s e^{-2s} + \frac{1}{2} \int e^{-2s} ds \\
&= -\frac{1}{2} s e^{-2s} + \frac{1}{2} (-\frac{1}{2}) e^{-2s} + C \\
&= -\frac{1}{2} s e^{-2s} - \frac{1}{4} e^{-2s} + C
\end{aligned} $$
(b) $\int \ln(x+1) dx$
Solution:
Let $u = \ln(x+1), \quad dv = dx$
$du = \frac{1}{x+1} dx, \quad v = x$
$$ \begin{aligned}
\int \ln(x+1) dx &= \ln(x+1) \cdot x - \int x \cdot \frac{1}{x+1} dx \\
&= x \ln(x+1) - \int \frac{x+1-1}{x+1} dx \\
&= x \ln(x+1) - \int (1 - \frac{1}{x+1}) dx \\
&= x \ln(x+1) - x + \ln|x+1| + C
\end{aligned} $$
(c) $\int t \sin 2t dt$
Solution:
Let $u = t, \quad dv = \sin 2t dt$
$du = dt, \quad v = -\frac{1}{2} \cos 2t$
$$ \begin{aligned}
\int t \sin 2t dt &= t(-\frac{1}{2} \cos 2t) - \int (-\frac{1}{2} \cos 2t) dt \\
&= -\frac{1}{2} t \cos 2t + \frac{1}{2} \int \cos 2t dt \\
&= -\frac{1}{2} t \cos 2t + \frac{1}{2} \cdot \frac{1}{2} \sin 2t + C \\
&= -\frac{1}{2} t \cos 2t + \frac{1}{4} \sin 2t + C
\end{aligned} $$
(d) $\int x 2^x dx$
Solution:
Let $u = x, \quad dv = 2^x dx$
$du = dx, \quad v = \frac{2^x}{\ln 2}$
$$ \begin{aligned}
\int x 2^x dx &= x \cdot \frac{2^x}{\ln 2} - \int \frac{2^x}{\ln 2} dx \\
&= \frac{x \cdot 2^x}{\ln 2} - \frac{1}{\ln 2} \int 2^x dx \\
&= \frac{x \cdot 2^x}{\ln 2} - \frac{1}{\ln 2} \cdot \frac{2^x}{\ln 2} + C \\
&= \frac{x \cdot 2^x}{\ln 2} - \frac{2^x}{(\ln 2)^2} + C
\end{aligned} $$
(e) $\int x \cos 5x dx$
Solution:
Let $u = x, \quad dv = \cos 5x dx$
$du = dx, \quad v = \frac{1}{5} \sin 5x$
$$ \begin{aligned}
\int x \cos 5x dx &= x \cdot \frac{1}{5} \sin 5x - \int \frac{1}{5} \sin 5x dx \\
&= \frac{1}{5} x \sin 5x - \frac{1}{5} \int \sin 5x dx \\
&= \frac{1}{5} x \sin 5x - \frac{1}{5} (-\frac{1}{5} \cos 5x) + C \\
&= \frac{1}{5} x \sin 5x + \frac{1}{25} \cos 5x + C
\end{aligned} $$
(f) $\int e^x \cos x dx$
Solution:
Let $u = \cos x, \quad dv = e^x dx$
$du = -\sin x dx, \quad v = e^x$
$$ \int e^x \cos x dx = \cos x \cdot e^x - \int e^x (-\sin x dx) $$
$$ \int e^x \cos x dx = \cos x \cdot e^x + \int e^x \sin x dx \quad (1) $$
For $\int e^x \sin x dx$,
Let $u = \sin x, \quad dv = e^x dx$
$du = \cos x dx, \quad v = e^x$
$$ \int e^x \sin x dx = \sin x \cdot e^x - \int e^x (\cos x dx) $$
By substituting $\int e^x \sin x dx = \sin x \cdot e^x - \int e^x \cos x dx$ in (1), we get
$$ \int e^x \cos x dx = \cos x \cdot e^x + \sin x \cdot e^x - \int e^x \cos x dx $$
$$ 2 \int e^x \cos x dx = \cos x \cdot e^x + \sin x \cdot e^x + C_1 $$
$$ \therefore \int e^x \cos x dx = \frac{1}{2} (\cos x \cdot e^x + \sin x \cdot e^x) + C, \text{ where } C = \frac{1}{2}C_1 $$