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## Monday, 4 August 2014

### CHAPTER 7- Some Standard Integrals

The following table  includes these general forms as “standard integral forms”.
 Integrals Substitution used Result 15.$\int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}\,\,dx}$ $x = a\sin \theta$ ${\sin ^{ - 1}}\dfrac{x}{a} + C$ 16. $\int {\dfrac{1}{{{a^2} + {x^2}}}\,\,dx}$ $x = a\tan \theta$ $\dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C$ 17. $\int {\dfrac{1}{{x\sqrt {{x^2} - {a^2}} }}\,\,dx}$ $x = a\sec \theta$ $\dfrac{1}{a}{\sec ^{ - 1}}\dfrac{x}{a} + C$ 18. $\int {\dfrac{1}{{{x^2} - {a^2}}}\,\,dx}$ $x = a\sec \theta$Alternatively theexpression can be splitin to separated fractions $\dfrac{1}{{2a}}\ln \left| {\dfrac{{x - a}}{{x + a}}} \right| + C$ 19. $\int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}\,\,dx}$ $x = a\tan \theta$ $\ln \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$ 20. $\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}\,\,dx}$ $x = a\sec \theta$ $\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$ *21. $\int {\sqrt {\dfrac{{a - x}}{{a + x}}} \,\,dx}$ $x = a\cos 2\theta$ $\sqrt {{a^2} - {x^2}} - a{\cos ^{ - 1}}\dfrac{x}{a} + C$ *22 $\int {\sqrt {\dfrac{{x - a}}{{b - x}}} \,\,dx}$ $x = a{\cos ^2}\theta + b{\sin ^2}\theta$ $(b - a)\left\{ {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{x}{a} - \dfrac{1}{{2a}}\sqrt {{a^2} - {x^2}} } \right\} + C$ * need not be memorized
We will extend this table further as we progress through the rest of this chapter.
As stated earlier, many integrals can be converted to these ‘standard’ forms and thats where lies the use in committing these results to memory.
We will now discuss in more detail how to convert a given integral into one of these standard forms, if possible.
Suppose that we have a quadratic expression $Q(x) = a{x^2} + bx + c$. It should be obvious that integrals of the form $\int {\dfrac{1}{{Q(x)}}} \,\,dx$ can be evaluated using either $(17)$ or $(21)$ {depending on whether $Q(x)$ takes the form of ${X^2} + {A^2}$ or ${A^2} - {X^2}$ upon rearrangement ; $X$ depends on $x$ linearly; $A$ is a constant}. Similarly, integrals of the form $\int {\dfrac{1}{{\sqrt {Q(x)} }}} \,\,dx$ can be evaluate using $(15)$$(22)$, or $(23)$ depending on what form $Q(x)$ takes upon rearrangement. Let us go through some examples related to these forms.