In this lesson, we will consider functions constructed from simpler functions, including logarithms, exponentials and trigonometric functions.
Here is a summary of the standard integrals we have established so far:
For some constant $C$C:
| $\int\ f'\left(x\right)\ f\left(x\right)^n\ dx$∫ f′(x) f(x)n dx | $=$= | $\frac{1}{n+1}f\left(x\right)^{n+1}+C$1n+1f(x)n+1+C, for $n\ne-1$n≠−1 |
| $\int e^x\ dx$∫ex dx | $=$= | $e^x+C$ex+C |
| $\int e^{ax+b}\ dx$∫eax+b dx | $=$= | $\frac{1}{a}e^{ax+b}+C$1aeax+b+C |
| $\int\frac{1}{x}\ dx$∫1x dx | $=$= | $\ln\left|x\right|+C$ln|x|+C |
| $\int\frac{f'\left(x\right)}{f\left(x\right)}\ dx$∫f′(x)f(x) dx | $=$= | $\ln\left|f\left(x\right)\right|+C$ln|f(x)|+C |
| $\int\sin x\ dx$∫sinx dx | $=$= | $-\cos x+C$−cosx+C |
| $\int\cos x\ dx$∫cosx dx | $=$= | $\sin x+C$sinx+C |
| $\int\sec^2x\ dx$∫sec2x dx | $=$= | $\tan x+C$tanx+C |
| $\int\sin\left(ax+b\right)\ dx$∫sin(ax+b) dx | $=$= | $-\frac{1}{a}\cos\left(ax+b\right)+C$−1acos(ax+b)+C |
| $\int\cos\left(ax+b\right)\ dx$∫cos(ax+b) dx | $=$= | $\frac{1}{a}\sin\left(ax+b\right)+C$1asin(ax+b)+C |
| $\int\sec^2\left(ax+b\right)\ dx$∫sec2(ax+b) dx | $=$= | $\frac{1}{a}\tan\left(ax+b\right)+C$1atan(ax+b)+C |
The more familiar we are with these integrals and their variations, that easier it will be to recognise which one is involved in any given question.
It is also useful to know how to quickly sketch and transform the basic curves for questions involving area between curves.
Remember that for integrals that result in logarithmic functions, the function being integrated is usually in fraction form.
Read the following worked examples carefully as there are different combinations of functions involved.
Worked examples
Example 1
Find the primitive $\int2\left(t+\sin t\right)\ dt$∫2(t+sint) dt.
Think: This expression involves the sum of a trigonometric function and a function in $t$t. We will need to consider each function separately and add their integrals together at the end.
Do: To begin, we can move the constant through the integral sign and write:
$2\int\left(t+\sin t\right)\ dt$2∫(t+sint) dt
This is the same as:
$2\left[\int t\ dt+\int\sin t\ dt\right]$2[∫t dt+∫sint dt]
So, the primitive is:
$t^2-2\cos t+C$t2−2cost+C
Example 2
Find the area enclosed between the parabola $y=x^2$y=x2 and the hyperbola $y=6-\frac{6}{x+1}$y=6−6x+1 in the first quadrant.
Think: Firstly, we need to find exactly where these two curves intersect each other, and to this end we equate both functions and solve for $x$x.
Do:
| $x^2$x2 | $=$= | $6-\frac{6}{x+1}$6−6x+1 |
| $x^2\left(x+1\right)$x2(x+1) | $=$= | $6\left(x+1\right)-6$6(x+1)−6 |
| $x^3+x^2$x3+x2 | $=$= | $6x$6x |
| $x\left(x^2+x-6\right)$x(x2+x−6) | $=$= | $0$0 |
| $x\left(x+3\right)\left(x-2\right)$x(x+3)(x−2) | $=$= | $0$0 |
| $\therefore$∴ $x$x | $=$= | $0,-3,2$0,−3,2 |
The two points of intersection that interest us are $x=0$x=0 and $x=2$x=2, as they both occur either on the boundary of or in the first quadrant. Hence, the two curves intersect at the origin and at the point $\left(2,4\right)$(2,4) as shown in the sketch below:
The area of interest, shaded in green, is evaluated as follows:
| $A$A | $=$= | $\int_0^2\left(6-\frac{6}{x+1}\right)\ dx-\int_0^2x^2\ dx$∫20(6−6x+1) dx−∫20x2 dx |
| $=$= | $\int_0^2\left(6-\frac{6}{x+1}-x^2\right)\ dx$∫20(6−6x+1−x2) dx | |
| $=$= | $\left[6x-6\ln\left(x+1\right)-\frac{x^3}{3}\right]_0^2$[6x−6ln(x+1)−x33]20 | |
| $=$= | $\left(12-6\ln3-\frac{8}{3}\right)-\left(0\right)$(12−6ln3−83)−(0) | |
| $\approx$≈ | $2.74166$2.74166 |
Hence, the area enclosed is approximately $2.742$2.742 units2.
Example 3
$\int_1^4\frac{dx}{2\sqrt{x}\left(\sqrt{x}+1\right)}$∫41dx2√x(√x+1)
Think: In the form given, this one doesn't quite seem to fit the general form $\int\frac{f'\left(x\right)}{f\left(x\right)}\ dx$∫f′(x)f(x) dx.
We can manipulate this fraction, however, by noticing that:
$\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}=\frac{1}{2\sqrt{x}}$ddx√x=12√x
Do: We can more easily see how to integrate this function by first lifting factor $\frac{1}{2\sqrt{x}}$12√x into the numerator:
| $\int_1^4\frac{dx}{2\sqrt{x}\left(\sqrt{x}+1\right)}$∫41dx2√x(√x+1) | $=$= | $\int_1^4\frac{\frac{1}{2\sqrt{x}}}{\sqrt{x}+1}\ dx$∫4112√x√x+1 dx |
| $=$= | $\left[\ln\left(\sqrt{x}+1\right)\right]_1^4$[ln(√x+1)]41 | |
| $=$= | $\ln3-\ln2$ln3−ln2 | |
| $=$= | $\ln1.5$ln1.5 |
Example 4
Evaluate $\int\frac{e^x-1}{2e^x-2x+1}\ dx$∫ex−12ex−2x+1 dx.
Think: Here, an exponential function is involved but the $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x) pattern is easy to spot.
Do: Integrating:
| $\int\frac{e^x-1}{2e^x-2x+1}\ dx$∫ex−12ex−2x+1 dx | $=$= | $\frac{1}{2}\int\frac{2e^x-2}{2e^x-2x+1}\ dx$12∫2ex−22ex−2x+1 dx |
| $=$= | $\frac{1}{2}\ln\left(2e^x-2x+1\right)+C$12ln(2ex−2x+1)+C |
Example 5
Find the area between the curves $y=e^x$y=ex and $y=2-e^x$y=2−ex and the line $x=1$x=1.
Think: A sketch is always good place to start for problems involving area.
Do: Note that curve given by $y=2-e^x$y=2−ex can be thought of as the curve $y=e^x$y=ex reflected across the $x$x-axis and vertically translated up by $2$2 units.
The area depicted in green is found as follows:
| $A$A | $=$= | $\int_0^1e^x\ dx-\int_0^12-e^x\ dx$∫10ex dx−∫102−ex dx |
| $=$= | $\int_0^1e^x-\left(2-e^x\right)\ dx$∫10ex−(2−ex) dx | |
| $=$= | $2\int_0^1e^x-1\ dx$2∫10ex−1 dx | |
| $=$= | $2\left[e^x-x\right]_0^1$2[ex−x]10 | |
| $=$= | $2\left[e^x-x\right]_0^1$2[ex−x]10 | |
| $=$= | $2\left(e-2\right)$2(e−2) | |
| $\approx$≈ | $1.4366$1.4366 |
Practice questions
Question 1
Determine $\int\frac{3\cos x}{5-\sin x}dx$∫3cosx5−sinxdx.
You may use $C$C to represent the constant of integration.
Question 2
Evaluate $\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\left(4\cos x+\cos4x\right)dx$∫π6−π6(4cosx+cos4x)dx.
Question 3
Find the value of $\int\frac{\sqrt{x}+3}{x}dx$∫√x+3xdx.
You may use $C$C to represent the constant of integration.
Question 4
Find the exact value of $\int_{\ln2}^{\ln6}\frac{e^{3x}+9}{e^x}dx$∫ln6ln2e3x+9exdx.