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Integration by substitution

Integration by substitution

The key idea behind integration by substitution is the following
$$
\int_a^b f(u(x)) \frac{du}{dx}dx = \int_{u(a)}^{u(b)}  f(u)du 
$$
The tricky part of such questions is knowing what substitution to make (that is what function $u(x)$ will substantially simplify your integral). The second issue is then suitably rearranging your limits

Integration by Substitution 1

Evaluate the following integral $$ \int \frac{2x}{\sqrt{x^2 + 12}}dx $$ using the substitution $u = x^2 + 12$
solution - press button to display
Differentiating the substitution gives us $\frac{du}{dx} = 2x$ We can now see that the integral can be rewritten as $$ \int \frac{2x}{\sqrt{x^2 + 12}}dx = \int \frac{1}{\sqrt{x^2 + 12}}2xdx = \int \frac{1}{\sqrt{u}}du $$ $$ \int \frac{1}{\sqrt{u}}du = 2u^\frac{1}{2}+ c = 2\sqrt{x^2+ 12} + c $$

Integration by Substitution 2

Evaluate the following integral $$ \int_0^1\sqrt{1 - x^2}dx $$ using the substitution $x = \sin \theta$
solution - press button to display
Differentiating the substitution gives us $\frac{dx}{d\theta} = \cos\theta$ For now, we shall ignore the limits. We can rewrite the integral (using the pythagorean identity and then the double angle formulae) $$ \begin{array}{rcl} \int \sqrt{1 - x^2}dx &=& \int \sqrt{1 - \sin^2\theta}\cos\theta d\theta \\ &=& \int\cos^2\theta d\theta \\ &=& \int\frac{1}{2} + \frac{1}{2}\cos(2\theta) d\theta \\ &=& \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + c \end{array} $$ Let us now consider the limits. If $x = 0$, then $\sin(\theta) = 0\Rightarrow \theta = 0$. If $x = 1$, then $\sin(\theta) = 1\Rightarrow \theta = \frac{\pi}{2}$ Therefore $$ \begin{array}{rcl} \int_0^1\sqrt{1 - x^2}dx &=& \left[\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta\right]_0^\frac{\pi}{2} \\ &=& \frac{\pi}{4} + \frac{1}{4}\sin\pi - \left(0 +\sin(0)\right) \\ &=& \frac{\pi}{4} \end{array} $$ (we can see immediately this is correct as the original curve encloses one quarter of the unit circle!)