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RuiAce

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Integration techniques - scaffold examples
« on: June 21, 2019, 07:03:48 pm »
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Integration in specialist maths builds directly onto those demonstrated in maths methods. We now have a larger range of functions that we may integrate, and also introduce integration techniques (much like differentiation techniques such as the product rule).

Check that after completion of Unit 4, you would not have trouble following all of the examples and brief explanations provided.

Integrating squares of trigonometric functions

\begin{align*}
\int \sin^2 3x\,dx &= \frac12 \int (1-\cos 6x)\,dx\\
&= \frac{x}{2} - \frac{\sin 6x}{12}+C
\end{align*}
In general, for integrals of the form \(\sin^2ax\) and \(\cos^2ax\), the double angle identity \(\cos 2x = 2\cos^2 x - 1 = 1-2\sin^2 x\) should be exploited.
\[ \int \sec^2 5x\,dx = \frac15 \tan 5x + C \]
The integral of secant-squared is a standard integral on your formula sheet!
\begin{align*}
\int \tan^2x\,dx &= \int \sec^2x - 1\,dx\\
&= \tan x - x + C
\end{align*}
As a special trick, when integrating tangent-squared, apply the Pythagorean identity beforehand.

Integration by substitution
For the integral \( \displaystyle\int_1^4 \frac{e^x}{\sqrt{x}} \), we may apply the substitution
\[ u = \sqrt{x} \implies du = \frac{1}{2\sqrt{x}}\,dx. \]
As this is a definite integral, we change the boundaries appropriate:
\begin{align*}
x=4&\implies u=2\\
x=1&\implies u=1
\end{align*}
Then we obtain
\begin{align*}
\int_1^4 \frac{e^x}{\sqrt{x}}\,dx &= 2\int_1^2 e^u\,du\\
&= 2[e^u]_1^2\\
&= 2(e^2-e)
\end{align*}
Whereas for indefinite integrals, we instead have to sub away \(u\) back for \(x\)!

Integrating the reciprocal function
\[ \int \frac{2}{3x+5}\,dx = \frac23 \ln |3x+5|+c \]
Unlike in methods, where we were more lenient, in spesh we understand that an absolute value must appear when we integrate back to log! This is to address the gap of negative values being denied in our final answer.

Use of inverse trigonometric functions
\[ \int \frac{1}{144+x^2}\,dx = \frac{1}{12} \tan^{-1} \frac{x}{12}+c \]
Remember that for integrals into arctangent, a factor of \( \frac{1}{a}\) comes out in front! This doesn't apply for integrals that go to arcsine instead!
\begin{align*}
\int_0^1 \frac{dx}{\sqrt{4-2x^2}} &= \int_0^1 \frac{dx}{\sqrt{2\left(2-x^2 \right)}}\\
&= \int_0^1 \frac{dx}{\sqrt{2}\sqrt{2-x^2}}\\
&= \frac{1}{\sqrt2} \int_0^1 \frac{dx}{\sqrt{2-x^2}}\\
&= \frac1{\sqrt2} \left[ \sin^{-1} \left( \frac{x}{\sqrt2} \right) \right]_0^{1}\\
&= \frac{1}{\sqrt2}\left( \sin^{-1} \frac1{\sqrt2} - \sin^{-1}0 \right)\\
&= \frac\pi{4\sqrt2}
\end{align*}
The integral formula only works when we can assume that the coefficient on \(x^2\) is \(1\)! When this is not the case, a bit of manipulation has to come first.

Simple partial fractions
If we wanted to consider, say, \( \displaystyle \int \frac{dx}{(x-1)(x-2)(x-3)} \)...
\[ \text{Write }\frac{1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \]
\[ \text{Then, }1 = A(x-2)(x-3)+B(x-1)(x-3) + C(x-1)(x-2) \]
Sub:
- \(x=1\) to obtain \(1 = 6A \implies A = \frac16\)
- \(x=2\) to obtain \(1 = -B\implies B = -1\)
- \(x=3\) to obtain \(1 = 6C \implies C = \frac16\)
\begin{align*}
\therefore \int \frac{dx}{(x-1)(x-2)(x-3)} &= \int \frac{1}{6(x-1)} - \frac{1}{x-2} + \frac{1}{6(x-3)}\,dx\\
&= \frac16 \ln |x-1| - \ln |x-2| + \frac16 \ln |x-3|+c\\
&= \frac16 \ln \left| \frac{(x-1)(x-3)}{(x-2)^6} \right|
\end{align*}
That last step wasn't really necessary. That's just to see how well versed you are at logarithm laws.

Integration by parts
\begin{align*}
\int x^2\cos x\,dx &= x^2\sin x - \int 2x\sin x\,dx\\
&= x^2\sin x - \left(-2x\cos x + \int 2\cos x\,dx \right)\\
&= x^2\sin x + 2x\cos x - 2\sin x + c
\end{align*}
Sometimes you need to apply integration by parts more than once, unlike with the product rule!

According to the rule of LIATE, the algebraic expression \(x^2\) should be differentiated, leaving the trigonometric expression \(\sin x\) to be integrated.
« Last Edit: June 21, 2019, 07:08:04 pm by RuiAce »