Question 13.4: Use Table 13.1 and the properties of a linear operator to in......
Use Table 13.1 and the properties of a linear operator to integrate the following expressions:
(a) x² + 9
(b)3t^{4}-{\sqrt{t}}
(c) \frac{1}{x}
(d) (t + 2)²
(e){\frac{1}{z}}+z
(f)4\mathrm{e}^{2z}
(g) 3 sin 4t
(h) 4 cos(9x + 2)
(i)3\mathrm{e}^{2z}
(j)\frac{\sin x+\cos x}{2}
(k) 2t − e^t
(l)\tan\left({\frac{z-1}{2}}\right)
(m)\mathrm{e}^{t}+\mathrm{e}^{-t}
(n) 3 sec(4x − 1)
(o) 2 cot 9x
(p) 7 cosec(π/3)
| Table 13.1 The integrals of some common functions. |
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| f(x) | \int f(x) d x | f(x) | \int f(x) \mathrm{d} x |
| k, constant | kx + c | \cos (a x+b) | \frac{\sin (a x+b)}{a}+c |
| x^n | \frac{x^{n+1}}{n+1}+c \quad n \neq-1 | \tan x | \ln |\sec x|+c |
| x^{-1}=\frac{1}{x} | \ln |x|+c | \tan a x | \frac{\ln |\sec a x|}{a}+c |
| \mathrm{e}^x | \mathrm{e}^x+c | \tan (a x+b) | \frac{\ln |\sec (a x+b)|}{a}+c |
| \mathrm{e}^{-x} | -\mathrm{e}^{-x}+c | \operatorname{cosec}(a x+b) | \frac{1}{a}\{\ln \mid \operatorname{cosec}(a x+b) |
| \mathrm{e}^{a x} | \frac{\mathrm{e}^{a x}}{a}+c | -\cot (a x+b) \mid\}+c | |
| \sin x | -\cos x+c | \sec (a x+b) | \frac{1}{a}\{\ln \mid \sec (a x+b) |
| \sin a x | \frac{-\cos a x}{a}+c | +\tan (a x+b) \mid\}+c | |
| \sin (a x+b) | \frac{-\cos (a x+b)}{a}+c | \cot (a x+b) | \frac{1}{a}\{\ln |\sin (a x+b)|\}+c |
| \cos x | \sin x+c | \frac{1}{\sqrt{a^2-x^2}} | \sin ^{-1} \frac{x}{a}+c |
| \cos a x | \frac{\sin a x}{a}+c | \frac{1}{a^2+x^2} | \frac{1}{a} \tan ^{-1} \frac{x}{a}+c |
(a) \int x^2+9 \mathrm{~d} x=\int x^2 \mathrm{~d} x+\int 9 \mathrm{~d} x \quad \text { using linearity }
=\frac{x^3}{3}+9 x+c \quad \text { using Table } 13.1
Note that only a single constant of integration is required.
(b) \int 3 t^4-\sqrt{t} \mathrm{~d} t=3 \int t^4 \mathrm{~d} t-\int t^{1 / 2} \mathrm{~d} t \quad \text { using linearity }
\begin{aligned}& =3\left(\frac{t^5}{5}\right)-\frac{t^{3 / 2}}{3 / 2}+c \quad \text { using Table } 13.1 \\& =\frac{3 t^5}{5}-\frac{2 t^{3 / 2}}{3}+c\end{aligned}
(c) \int \frac{1}{x} \mathrm{~d} x=\ln |x|+c \quad \text { using Table 13.1. }
Sometimes it is convenient to use the laws of logarithms to rewrite answers involving logarithms. For example, we can write \ln |x|+c \text { as } \ln |x|+\ln |A| \text { where } c=\ln |A| \text {. } This enables us to write the integral as
\int \frac{1}{x} \mathrm{~d} x=\ln |A x|
(d) \int(t+2)^2 \mathrm{~d} t=\int t^2+4 t+4 \mathrm{~d} t=\frac{t^3}{3}+2 t^2+4 t+c
(e) \int \frac{1}{z}+z \mathrm{~d} z=\ln |z|+\frac{z^2}{2}+c
(f) \int 4 \mathrm{e}^{2 z} \mathrm{~d} z=\frac{4 \mathrm{e}^{2 z}}{2}+c=2 \mathrm{e}^{2 z}+c
(g) \int 3 \sin (4 t) \mathrm{d} t=-\frac{3 \cos 4 t}{4}+c
(h) \int 4 \cos (9 x+2) d x=\frac{4 \sin (9 x+2)}{9}+c
(i) \int 3 \mathrm{e}^{2 z} \mathrm{~d} z=\frac{3 \mathrm{e}^{2 z}}{2}+c
(j) \int \frac{\sin x+\cos x}{2} \mathrm{~d} x=\frac{-\cos x+\sin x}{2}+c
(k) \int 2 t-\mathrm{e}^t \mathrm{~d} t=t^2-\mathrm{e}^t+c
(l) \int \tan \left(\frac{z-1}{2}\right) \mathrm{d} z=2 \ln \left|\sec \left(\frac{z-1}{2}\right)\right|+c
(m) \int \mathrm{e}^t+\mathrm{e}^{-t} \mathrm{~d} t=\mathrm{e}^t-\mathrm{e}^{-t}+c
(n) \int 3 \sec (4 x-1) \mathrm{d} x=\frac{3}{4} \ln |\sec (4 x-1)+\tan (4 x-1)|+c
(o) \int 2 \cot 9 x d x=\frac{2}{9} \ln |\sin 9 x|+c
(p) \int 7 \operatorname{cosec}(\pi / 3) \mathrm{d} x=\{7 \operatorname{cosec}(\pi / 3)\} x+c \quad \text { as } \operatorname{cosec}(\pi / 3) \text { is a constant }