Integrating exp, log, trig and other functions

At the basic level, we saw how to integrate polynomials – further integration includes integrating exp/recip/trig and other functions. It is possible to differentiate all of these functions and, similarly, we may integrate them too. Recall from the fundamental theorem of calculus that differentiation and integration can be thought of as reverse processes to one other. As well as differentiating polynomials, we know how to differentiate various exponential, logarithmic and trigonometric functions:


Hence, we can find the integrals of such derivatives by going backwards. However, when we integrate we must remember the constant of integration:

\begin{array}{lll}\int e^x dx&=&e^x+c\\\int\frac{1}{x}dx&=&\ln(x)+c\\\int\cos(x)dx&=&\sin(x)+c\\\int\sin(x)dx&=&-\cos(x)+c\end{array}

Note that when differentiating and integrating trigonometric functions, we must work in radians. A question may expect you to integrate linear combinations of the above functions. See Example 1. In addition, we can apply this type of integration in questions similar to those we learned to answer at the basic level (see basic integration). This may include finding f(x) from f'(x) and use a point to find the integration constant. See Example 2.  

In addition to the above, we also have, for example, \int\cos(kx)dx=\frac{1}{k}\sin(kx)+c, for any constant k. We can see this from inspection, that is, differentiating \sin(kx) (using the chain rule if necessary). The derivative is k\cos(kx) and so \int k\cos(kx)dx=\sin(kx)+c. By adjusting the factors we can conclude that \int \cos(kx)dx=\frac{1}{k}\sin(kx)+c. We can apply a similar process to the other integrals above. Alternatively, we integrate these integrals with k in the argument using integration by substitution.

Using the Formula Booklet

You may recall from differentiating trigonometric functions that the formula booklet tells us how to differentiate \tan(kx), \sec(kx), \cot(kx) and \text{cosec}(kx). The formula booklet also tells us how to integrate these functions and also tells us how to integrate \sec^2(kx) as follows:


For example, the integral of \cot(4x) is \frac{1}{4}\ln\vert\sin(4x)\vert +c. We can check this by differentiating \frac{1}{4}\ln\vert\sin(4x)\vert +c using the chain rule. See Example 3 for another example using the above in a definite integral. The formula booklet also provides alternative answers for the last two integrals – see the formula booklet. It is possible that we must manipulate an integral before we can use one of the above integrals (see Example 2 in integrating with trigonometric identities).


Evaluate \int\left(4\sin(x)-\frac{1}{3}\cos(x)-\frac{5-3xe^x}{2x}\right)dx.


The first two terms in the integrand are just multiples of trigonometric functions. However, we must rewrite the fraction as \frac{5-3xe^x}{2x}=\frac{5}{2x}-\frac{3}{2}e^x. It follows that


The curve of y=f(x) passes through the point \left(\pi,7\right). Given that f'(x)=-3\sin(x), find an expression for f(x) and sketch its curve.


To find an expression for f(x) we must integrate f'(x). It follows that f(x)=3\cos(x)+c. The coordinates of the point allows us to find the integration constant. That is, 7=3\cos(\pi)+c=-3+c. Hence, c=10 and the expression for f(x)=3\cos(x)+10.

further integration

Evaluate \int_{0}^{\frac{\pi}{12}}\left(3\sec^2(2x)-2\tan(3x)\right)dx exactly.


\begin{array}{lll}\int_{0}^{\frac{\pi}{12}}\left(3\sec^2(2x)-2\tan(3x)\right)dx&=&[\frac{3}{2}\tan(2x)-\frac{2}{3}\ln\vert \sec(3x)\vert]_{0}^{\frac{\pi}{12}}\\&=&\left(\frac{3}{2}\tan\left(\frac{2\pi}{12}\right)-\frac{2}{3}\ln\vert \sec\left(\frac{3\pi}{12}\right)\right)-\left(\frac{3}{2}\tan\left(0\right)-\frac{2}{3}\ln\vert \sec\left(0\right)\vert\right)\\&=&\frac{\sqrt{3}}{2}-\frac{2}{3}\ln(\sqrt{2})\\&=&\frac{\sqrt{3}}{2}-\frac{1}{3}\ln(2)\end{array}