Trigonometric Substitution Integration Technique
Smaridasa: The next integration technique we will learn is the Trigonometric Substitution Integration Technique. This technique is usually used when the "Algebraic Substitution Integration" technique or the "integration by parts" technique cannot solve the integral problem. The Trigonometric Substitution Integration Technique is specifically used if there are forms like \(\sqrt{a^2 - b^2 x^2}\), \(\sqrt{a^2 + b^2 x^2}\), and \(\sqrt{a^2 x^2 - b^2}\). For more details, let's look at the following explanation.
Trigonometric Substitution Integration Technique
Trigonometric Identities and Inverse Trigonometric Functions
1) Trigonometric Identities:
Actually, the purpose of this trigonometric substitution technique is to direct the problem into the form of trigonometric identity equations, namely:
\[
\sin^2 t + \cos^2 t = 1,
\quad 1 + \tan^2 t = \sec^2 t,
\quad 1 + \cot^2 t = \csc^2 t.
\]
2) Inverse Trigonometric Functions:
The following are the inverse forms:
\[
\text{If } \sin t = f(x),\ \text{then } t = \arcsin(f(x))
\]
\[
\text{If } \cos t = f(x),\ \text{then } t = \arccos(f(x))
\]
\[
\text{If } \tan t = f(x),\ \text{then } t = \arctan(f(x))
\]
\[
\text{If } \cot t = f(x),\ \text{then } t = \text{arccot}(f(x))
\]
\[
\text{If } \sec t = f(x),\ \text{then } t = \text{arcsec}(f(x))
\]
\[
\text{If } \csc t = f(x),\ \text{then } t = \text{arccsc}(f(x))
\]
Example of Inverse Trigonometric Functions:
1) Determine the inverse of:
a) \(\sin t = \frac{1}{2}\)
b) \(\cos t = 3x\)
c) \(\tan t = \frac{x-2}{5}\)
Smaridasa:
a) \(\sin t = \frac{1}{2} \ \rightarrow\ t = \arcsin\frac{1}{2} \ \rightarrow\ t = 30^\circ\)
b) \(\cos t = 3x \ \rightarrow\ t = \arccos(3x)\)
c) \(\tan t = \frac{x-2}{5} \ \rightarrow\ t = \arctan\left(\frac{x-2}{5}\right)\)
Forms of Trigonometric Substitution
The following are the substitution forms we will use:
a) For the form \(\sqrt{a^2 - b^2 x^2}\), substitute \(x = \frac{a}{b} \sin t\) or \(x = \frac{a}{b} \cos t\).
b) For the form \(\sqrt{a^2 + b^2 x^2}\), substitute \(x = \frac{a}{b} \tan t\) or \(x = \frac{a}{b} \cot t\).
c) For the form \(\sqrt{b^2 x^2 - a^2}\), substitute \(x = \frac{a}{b} \sec t\) or \(x = \frac{a}{b} \csc t\).
Example Problems:
2) Determine the integral result of \(\int \frac{x^2}{\sqrt{1-x^2}} \, dx\)?
Smaridasa:
*) This problem is difficult to solve with algebraic substitution or integration by parts, so we solve it using the trigonometric substitution technique.
*) For the form \(\sqrt{1-x^2}\), substitute \(x = \sin t\) or \(x = \cos t\).
*) First, substitute with \(x = \sin t\):
\(x = \sin t \ \rightarrow\ t = \arcsin x\)
\(x = \sin t \ \rightarrow\ \frac{dx}{dt} = \cos t \ \rightarrow\ dx = \cos t \, dt\)
\(\sqrt{1-x^2} = \sqrt{1-(\sin t)^2} = \sqrt{\cos^2 t} = \cos t\).
Use the formula: \(\sin^2 t = \frac{1}{2} - \frac{1}{2}\cos 2t\), and \(\sin 2t = 2\sin t \cos t\).
Also \(\cos t = \sqrt{1-\sin^2 t} = \sqrt{1-x^2}\).
Returning to the problem, we replace all \(x\) and \(dx\) variables:
\[
\int \frac{x^2}{\sqrt{1-x^2}} \, dx = \int \frac{(\sin t)^2}{\cos t} \cdot \cos t \, dt
= \int (\sin t)^2 \, dt
= \int \left( \frac{1}{2} - \frac{1}{2}\cos 2t \right) dt
\]
\[
= \frac{1}{2}t - \frac{1}{4}\sin 2t + c
= \frac{1}{2}t - \frac{1}{2}\sin t \cos t + c
\]
(Change back to \(x\) form)
\[
= \frac{1}{2} \arcsin x - \frac{1}{2}x\sqrt{1-x^2} + c
\]
\[
= \frac{1}{2} \arcsin x - \frac{x}{2}\sqrt{1-x^2} + c
\]
So, the result is:
\[
\int \frac{x^2}{\sqrt{1-x^2}} \, dx = \frac{1}{2} \arcsin x - \frac{x}{2}\sqrt{1-x^2} + c.
\]
*) Second, substitute with \(x = \cos t\):
\(x = \cos t \ \rightarrow\ t = \arccos x\)
\(x = \cos t \ \rightarrow\ \frac{dx}{dt} = -\sin t \ \rightarrow\ dx = -\sin t \, dt\)
\(\sqrt{1-x^2} = \sqrt{1-(\cos t)^2} = \sqrt{\sin^2 t} = \sin t\).
Use the formula: \(\cos^2 t = \frac{1}{2} + \frac{1}{2}\cos 2t\), and \(\sin 2t = 2\sin t \cos t\).
Also \(\sin t = \sqrt{1-\cos^2 t} = \sqrt{1-x^2}\).
Returning to the problem, we replace all \(x\) and \(dx\) variables:
\[
\int \frac{x^2}{\sqrt{1-x^2}} \, dx = \int \frac{(\cos t)^2}{\sin t} \cdot (-\sin t) \, dt
= -\int (\cos t)^2 \, dt
= -\int \left( \frac{1}{2} + \frac{1}{2}\cos 2t \right) dt
\]
\[
= -\left( \frac{1}{2}t + \frac{1}{4}\sin 2t \right) + c
= -\frac{1}{2}t - \frac{1}{2}\sin t \cos t + c
\]
(Change back to \(x\) form)
\[
= -\frac{1}{2} \arccos x - \frac{1}{2}x\sqrt{1-x^2} + c
\]
\[
= -\frac{1}{2} \arccos x - \frac{x}{2}\sqrt{1-x^2} + c
\]
So, the result is:
\[
\int \frac{x^2}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \arccos x - \frac{x}{2}\sqrt{1-x^2} + c.
\]
Both assumptions above give different results, but both integral results are correct. If the problem has multiple choices, only one will certainly be present, not both. And if the problem is in essay form, the result depends on the trigonometric substitution we choose, and both are correct.
3) Determine the integral result of \(\int \frac{x^2}{\sqrt{4-9x^2}} \, dx\)?
Smaridasa:
*) The form is \(\sqrt{4-9x^2} = \sqrt{2^2 - 3^2 x^2}\), substitute \(x = \frac{2}{3} \sin t\).
\(x = \frac{2}{3} \sin t \ \rightarrow\ \sin t = \frac{3x}{2} \ \rightarrow\ t = \arcsin \left( \frac{3x}{2} \right)\)
\(x = \frac{2}{3} \sin t \ \rightarrow\ \frac{dx}{dt} = \frac{2}{3} \cos t \ \rightarrow\ dx = \frac{2}{3} \cos t \, dt\).
Use the formula: \(\sin^2 t = \frac{1}{2} - \frac{1}{2}\cos 2t\), and \(\sin 2t = 2\sin t \cos t\).
Also \(\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - \left( \frac{3x}{2} \right)^2} = \sqrt{1 - \frac{9}{4} x^2}\).
Returning to the problem, we replace all \(x\) and \(dx\) variables:
\[
\int \frac{x^2}{\sqrt{4-9x^2}} \, dx = \int \frac{\left( \frac{2}{3} \sin t \right)^2}{2 \cos t} \cdot \frac{2}{3} \cos t \, dt
= \int \frac{\frac{4}{9} \sin^2 t}{2 \cos t} \cdot \frac{2}{3} \cos t \, dt
= \frac{4}{27} \int \sin^2 t \, dt
\]
\[
= \frac{4}{27} \int \left( \frac{1}{2} - \frac{1}{2}\cos 2t \right) dt
= \frac{4}{27} \left( \frac{1}{2}t - \frac{1}{4}\sin 2t \right) + c
= \frac{4}{27} \left( \frac{1}{2}t - \frac{1}{2}\sin t \cos t \right) + c
\]
(Change back to \(x\) form)
\[
= \frac{4}{27} \left( \frac{1}{2} \arcsin \left( \frac{3x}{2} \right) - \frac{1}{2} \cdot \frac{3x}{2} \sqrt{1 - \frac{9}{4}x^2} \right) + c
\]
\[
= \frac{2}{27} \arcsin \left( \frac{3x}{2} \right) - \frac{x}{9} \sqrt{4-9x^2} + c
\]
So, the result is:
\[
\int \frac{x^2}{\sqrt{4-9x^2}} \, dx = \frac{2}{27} \arcsin \left( \frac{3x}{2} \right) - \frac{x}{9} \sqrt{4-9x^2} + c.
\]
4) Determine the integral result of \(\int \frac{1}{4+x^2} \, dx\)?
Smaridasa:
*) The form is \(4+x^2\), substitute \(x = 2\tan t\).
\(x = 2\tan t \ \rightarrow\ \tan t = \frac{x}{2} \ \rightarrow\ t = \arctan \left( \frac{x}{2} \right)\)
\(x = 2\tan t \ \rightarrow\ \frac{dx}{dt} = 2\sec^2 t \ \rightarrow\ dx = 2\sec^2 t \, dt\).
\[
4+x^2 = 4 + (2\tan t)^2 = 4 + 4\tan^2 t = 4(1 + \tan^2 t) = 4\sec^2 t
\]
Returning to the problem, we replace all \(x\) and \(dx\) variables:
\[
\int \frac{1}{4+x^2} \, dx = \int \frac{1}{4\sec^2 t} \cdot 2\sec^2 t \, dt = \int \frac{1}{2} \, dt = \frac{1}{2} t + c
\]
\[
= \frac{1}{2} \arctan \left( \frac{x}{2} \right) + c
\]
So, the result is:
\[
\int \frac{1}{4+x^2} \, dx = \frac{1}{2} \arctan \left( \frac{x}{2} \right) + c.
\]
5) Determine the integral result \(\int \sqrt{8+2x-x^2} \, dx\)?
Smaridasa:
*) Change the function into a perfect square form.
\[
8+2x-x^2 = 9-1+2x-x^2 = 9 - (1-2x+x^2) = 9 - (x-1)^2
\]
*) The form is \(9 - (x-1)^2\), substitute \(x-1 = 3\sin t\).
\(x-1 = 3\sin t \ \rightarrow\ \sin t = \frac{x-1}{3} \ \rightarrow\ t = \arcsin\left(\frac{x-1}{3}\right)\)
\(x-1 = 3\sin t \ \rightarrow\ x = 3\sin t + 1 \ \rightarrow\ \frac{dx}{dt} = 3\cos t \ \rightarrow\ dx = 3\cos t \, dt\).
\[
\sqrt{8+2x-x^2} = \sqrt{9 - (3\sin t)^2} = \sqrt{9 - 9\sin^2 t} = \sqrt{9(1 - \sin^2 t)} = \sqrt{9\cos^2 t} = 3\cos t
\]
Use the formula: \(\cos^2 t = \frac{1}{2} + \frac{1}{2}\cos 2t\), and \(\sin 2t = 2\sin t \cos t\).
Also \(\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - \left( \frac{x-1}{3} \right)^2} = \sqrt{1 - \frac{1}{9}(x-1)^2}\).
Returning to the problem, we replace all \(x\) and \(dx\) variables:
\[
\int \sqrt{8+2x-x^2} \, dx = \int 3\cos t \cdot 3\cos t \, dt = 9 \int \cos^2 t \, dt
\]
\[
= 9 \int \left( \frac{1}{2} + \frac{1}{2}\cos 2t \right) dt = 9 \left( \frac{1}{2}t + \frac{1}{4}\sin 2t \right) + c
\]
\[
= \frac{9}{2}t + \frac{9}{4} \cdot 2\sin t \cos t + c = \frac{9}{2}t + \frac{9}{2} \sin t \cos t + c
\]
(Change back to \(x\) form)
\[
= \frac{9}{2} \arcsin\left(\frac{x-1}{3}\right) + \frac{9}{2} \cdot \frac{x-1}{3} \cdot \sqrt{1 - \frac{1}{9}(x-1)^2} + c
\]
\[
= \frac{9}{2} \arcsin\left(\frac{x-1}{3}\right) + \frac{3}{2}(x-1) \sqrt{1 - \frac{1}{9}(x-1)^2} + c
\]
So, the result is:
\[
\int \sqrt{8+2x-x^2} \, dx = \frac{9}{2} \arcsin\left(\frac{x-1}{3}\right) + \frac{3}{2}(x-1) \sqrt{1 - \frac{1}{9}(x-1)^2} + c.
\]
6) Determine the integral result \(\int \frac{1}{x^2 + 2x + 5} \, dx\)?
Smaridasa:
*) Change the function into a perfect square form.
\[
x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 4
\]
*) The form is \((x+1)^2 + 4\), substitute \(x+1 = 2\tan t\).
\(x+1 = 2\tan t \ \rightarrow\ \tan t = \frac{x+1}{2} \ \rightarrow\ t = \arctan\left(\frac{x+1}{2}\right)\)
\(x+1 = 2\tan t \ \rightarrow\ x = 2\tan t - 1 \ \rightarrow\ \frac{dx}{dt} = 2\sec^2 t \ \rightarrow\ dx = 2\sec^2 t \, dt\).
\[
(x+1)^2 + 4 = (2\tan t)^2 + 4 = 4\tan^2 t + 4 = 4(1+\tan^2 t) = 4\sec^2 t
\]
Returning to the problem, we replace all \(x\) and \(dx\) variables:
\[
\int \frac{1}{x^2 + 2x + 5} \, dx = \int \frac{1}{4\sec^2 t} \cdot 2\sec^2 t \, dt = \int \frac{1}{2} \, dt = \frac{1}{2} t + c
\]
\[
= \frac{1}{2} \arctan \left( \frac{x+1}{2} \right) + c
\]
So, the result is:
\[
\int \frac{1}{x^2 + 2x + 5} \, dx = \frac{1}{2} \arctan \left( \frac{x+1}{2} \right) + c.
\]