- Trigonometric Integration

Even powers of \(\sin ⁡x\) and \(\cos ⁡x\)

LESSON 1
Determine
\[\int \sin^4⁡x\:dx\]
SOLUTION
  • Rewrite as multiples of \(\sin^2⁡x\)
\(\int \sin^2⁡x \sin^2⁡x \:dx\)
  • Use the identity
\(\displaystyle \sin^2⁡x={1-\cos ⁡2x\over2}\)
Therefore, we now have
\(\displaystyle \sin^4⁡x\:dx=\int \left({1\over 2}-{\cos⁡2x\over2}\right)\left({1\over2}-\cos{⁡2x\over2}\right)\:dx\)
\(\displaystyle =\int\left({1\over4}-\cos⁡{2x\over2}+\cos^2⁡{2x\over4}\right)\:dx\)
\(\displaystyle ={1\over4}\int(1-2 \cos⁡2x+\cos^2⁡2x)\: dx\)
\(\displaystyle ={1\over4} \left[\int1\: dx-2\int\cos⁡2x\:dx+\int\cos^2⁡2x\: dx\right]\)
\(\displaystyle ={1\over4} \left[x-2\left({1\over2} \sin⁡2x\right )+\int \cos^2⁡2x \: dx\right]\)

  • To integrate \(\cos^2⁡2x\) use the double angle formula
\(\cos⁡2x=2 \cos^2⁡x-1\)
\(⟹\cos⁡(2(2x))=2 \cos^2⁡(2x)-1\)
\(⟹\cos⁡4x=2 \cos^2⁡2x-1\)
\(\displaystyle ⟹\cos^2⁡2x={\cos⁡4x+1\over2}\)
\(\displaystyle ⟹\int \cos^2⁡2x \:dx={1\over2} \int (\cos⁡4x+1) \:dx\)
\(\displaystyle ={1\over2} \left[{1\over4} \sin⁡4x+x\right]\)
\(\displaystyle ={1\over8} \sin⁡4x+{x\over2}\)
  • Resubstituting
\(\displaystyle \int \sin^4⁡x \:dx={1\over4} \left[x-\sin⁡2x+\left({\sin⁡4x\over8}+{x\over2}\right)\right]+c\)
\(\displaystyle ={x\over4}-{\sin⁡2x\over4}+{\sin⁡4x\over32}+{x\over8}+c\)
\(\displaystyle ={3x\over8}-{\sin⁡2x\over4}+{\sin⁡4x\over32}+c\)
\(\displaystyle ={3x\over8}-{\sin⁡2x\over4}+{\sin⁡4x\over32}+c\)

In the previous example, we saw how to integrate even multiples of cosine but let’s look at it once again.
LESSON 2
Determine
\[\int \cos^2⁡x \:dx\]
SOLUTION
\(\int \cos^2⁡x \:dx\)
  • Use the identity
\(\cos⁡2x=2 \cos^2⁡x-1\)
\(\displaystyle \cos^2⁡x={1\over2}+{\cos⁡2x\over2}\)
\(\displaystyle \int \cos^2⁡x \:dx=\int \left({1\over2}+{\cos⁡2x\over2}\right)\: dx\)
\(\displaystyle ={x\over2}+{\sin⁡2x\over4}+c\)

Odd powers on \(\sin⁡ x\) and \(\cos ⁡x\)

LESSON 1
Determine
\[\int \sin^3⁡x \:dx\]
SOLUTION
  • We will rewrite \(\int \sin^n⁡x\:dx\) in the form \(\int \sin ⁡x (\sin^{n-1}⁡x )\:dx\) where \(n\) is odd.
\(\int \sin^3⁡x\:dx=\int \sin ⁡x \sin^2⁡x \:dx\)
  • Since \(\sin^2⁡x=1-\cos^2⁡x\)
\(\int \sin ⁡x \sin^2⁡x \:dx=\int \sin ⁡x(1-\cos^2⁡x) \:dx\)
\(=\int \sin ⁡x\:dx-\int \sin ⁡x \cos^2⁡x\:dx\)
  • We now seek to use \(\displaystyle \int f'(x)[f(x)]^n\:dx={[f(x)]^{n+1}\over n+1}+c\)
\(\int \sin⁡ x \sin^2⁡x\:dx=\int \sin⁡ x \:dx+\int (-\sin ⁡x) \cos^2⁡x\:dx\)
\(f(x)=\cos ⁡x→f'(x)=-\sin ⁡x\)
\(\displaystyle \int \sin ⁡x \sin^2⁡x\:dx=-\cos ⁡x+{\cos^3⁡x\over3}+c\)

LESSON 2
Determine
\[\int \cos^3⁡x \:dx\]
SOLUTION
  • Rewrite \(\int \cos^n⁡x\:dx\) in the form \(\int \cos ⁡x \cos^2⁡x\: dx\) where \(n\) is odd.
\(\int \cos^3⁡x \:dx=\int \cos⁡ x \cos^2⁡x\:dx\)
\(=\int \cos⁡ x (1-\sin^2⁡x)\:dx\)
\(=\int \cos ⁡x\:dx-\int \cos ⁡x \sin^2⁡x \:dx\)
\(\displaystyle =\sin ⁡x-{\sin^3⁡x\over3}+c\)

LESSON 3
Determine
\[\int \cos^3⁡x \sin^4⁡x\:dx\]
SOLUTION
  • Though we have a combination of odd and even powers we will simply manipulate the odd power.
\(\int \cos^3⁡x \sin^4⁡x\:dx=\int \cos ⁡x \cos^2⁡x \sin^4⁡x\:dx\)
\(=\int \cos ⁡x (1-\sin^2⁡x) \sin^4⁡x\:dx\)
\(=\int \cos ⁡x \sin^4⁡x-\int \cos ⁡x \sin^6⁡x \:dx\)
\(\displaystyle ={\sin^5⁡x\over5}-{\sin^7⁡x\over7}+c\)

Even powers of \(\tan⁡ x\)

LESSON 1
Determine
\[\int \tan^2⁡x\:dx\]
SOLUTION
\(\int \tan^2⁡x\:dx=\int \sec^2⁡x-1\:dx\)
  • Since \(\sec^2⁡x=1+\tan^2⁡x\)
\(\int \tan^2⁡x\:dx=\tan ⁡x-x+c\)