Week 5 Discussion Notes
Table of Contents
Trig Substitution
Example 1.
Calculate ∫ d x x 2 + 4 \displaystyle \int \frac{\diff{x}}{x^2 + 4} ∫ x 2 + 4 d x
Solution.
In general, for trig substitution problems, your goal is to get rid of everything next to the x 2 x^2 x 2 x x x x 2 + 4 x^2 + 4 x 2 + 4 tan 2 θ + 1 = sec 2 θ \tan^2\theta + 1 = \sec^2\theta tan 2 θ + 1 = sec 2 θ 4 tan 2 θ + 4 = 4 sec 2 θ 4\tan^2\theta + 4 = 4\sec^2\theta 4 tan 2 θ + 4 = 4 sec 2 θ
x = 2 tan θ ⟹ d x = 2 sec 2 θ d θ . x = 2\tan\theta \implies \diff{x} = 2\sec^2\theta \,\diff\theta. x = 2 tan θ ⟹ d x = 2 sec 2 θ d θ . Also, x 2 + 4 = 4 sec 2 θ x^2 + 4 = 4\sec^2\theta x 2 + 4 = 4 sec 2 θ
∫ d x x 2 + 4 = ∫ 2 sec 2 θ 4 sec 2 θ d θ = ∫ 1 2 d θ = 1 2 θ + C . \int \frac{\diff{x}}{x^2 + 4}
= \int \frac{2\sec^2\theta}{4\sec^2\theta} \,\diff\theta
= \int \frac{1}{2} \,\diff\theta
= \frac{1}{2} \theta + C. ∫ x 2 + 4 d x = ∫ 4 sec 2 θ 2 sec 2 θ d θ = ∫ 2 1 d θ = 2 1 θ + C . All that's left to do is turn everything back into terms of x x x x = 2 tan θ x = 2\tan\theta x = 2 tan θ θ = arctan x 2 \theta = \arctan\frac{x}{2} θ = arctan 2 x
1 2 arctan x 2 + C . \boxed{\frac{1}{2}\arctan\frac{x}{2} + C}. 2 1 arctan 2 x + C .
Example 2.
Calculate ∫ d x x 2 + 1 \displaystyle \int \frac{\diff{x}}{\sqrt{x^2 + 1}} ∫ x 2 + 1 d x
Solution.
Like before, we want to turn the x 2 + 1 x^2 + 1 x 2 + 1 x = tan θ x = \tan\theta x = tan θ
d x = sec 2 θ d θ and x 2 + 1 = tan 2 θ + 1 = sec 2 θ , \diff{x} = \sec^2\theta \,\diff\theta
\quad\text{and}\quad
x^2 + 1 = \tan^2\theta + 1 = \sec^2\theta, d x = sec 2 θ d θ and x 2 + 1 = tan 2 θ + 1 = sec 2 θ , so the integral becomes
∫ d x x 2 + 1 = ∫ sec 2 θ sec θ d θ = ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C . \int \frac{\diff{x}}{\sqrt{x^2 + 1}}
= \int \frac{\sec^2\theta}{\sec\theta} \,\diff\theta
= \int \sec\theta \,\diff\theta
= \ln\abs{\sec\theta + \tan\theta} + C. ∫ x 2 + 1 d x = ∫ sec θ sec 2 θ d θ = ∫ sec θ d θ = ln ∣ sec θ + tan θ ∣ + C . We have to do a little bit more work this time to get everything back into terms of x x x x = tan θ x = \tan\theta x = tan θ sec θ \sec\theta sec θ
x 2 + 1 = sec 2 θ ⟹ sec θ = x 2 + 1 , x^2 + 1 = \sec^2\theta \implies \sec\theta = \sqrt{x^2 + 1}, x 2 + 1 = sec 2 θ ⟹ sec θ = x 2 + 1 , so in the end, we get
ln ∣ sec θ + tan θ ∣ + C = ln ∣ x 2 + 1 + x ∣ + C . \ln\abs{\sec\theta + \tan\theta} + C
= \boxed{\ln\abs{\sqrt{x^2 + 1} + x} + C}. ln ∣ sec θ + tan θ ∣ + C = ln ∣ ∣ x 2 + 1 + x ∣ ∣ + C .
Example 3.
Calculate ∫ d x x 2 x 2 − 9 \displaystyle \int \frac{\diff{x}}{x^2\sqrt{x^2 - 9}} ∫ x 2 x 2 − 9 d x
Solution.
Same idea as before, but this time x = 3 tan θ x = 3\tan\theta x = 3 tan θ not work. Instead, if you notice that sec 2 θ − 1 = tan 2 θ \sec^2\theta - 1 = \tan^2\theta sec 2 θ − 1 = tan 2 θ x = 3 sec θ x = 3\sec\theta x = 3 sec θ
d x = 3 sec θ tan θ d θ and x 2 − 9 = 9 sec 2 θ − 9 = 9 tan 2 θ \diff{x} = 3\sec\theta\tan\theta \,\diff\theta
\quad\text{and}\quad
x^2 - 9 = 9\sec^2\theta - 9 = 9\tan^2\theta d x = 3 sec θ tan θ d θ and x 2 − 9 = 9 sec 2 θ − 9 = 9 tan 2 θ and
∫ d x x 2 x 2 − 9 = ∫ 3 sec θ tan θ 9 sec 2 θ ⋅ 3 tan θ d θ = 1 9 ∫ d θ sec θ = 1 9 ∫ cos θ d θ = 1 9 sin θ + C . \begin{aligned}
\int \frac{\diff{x}}{x^2\sqrt{x^2 - 9}}
= \int \frac{3\sec\theta\tan\theta}{9\sec^2\theta \cdot 3\tan\theta} \,\diff\theta
&= \frac{1}{9} \int \frac{\diff\theta}{\sec\theta} \\
&= \frac{1}{9} \int \cos\theta \,\diff\theta \\
&= \frac{1}{9} \sin\theta + C.
\end{aligned} ∫ x 2 x 2 − 9 d x = ∫ 9 sec 2 θ ⋅ 3 tan θ 3 sec θ tan θ d θ = 9 1 ∫ sec θ d θ = 9 1 ∫ cos θ d θ = 9 1 sin θ + C . From our substitution,
x = 3 sec θ ⟹ x 3 = 1 cos θ ⟹ cos θ = 3 x ⟹ cos 2 θ = 9 x 2 ⟹ sin 2 θ = 1 − cos 2 θ = 1 − 9 x 2 = x 2 − 9 x 2 ⟹ sin θ = x 2 − 9 x . \begin{aligned}
x = 3\sec\theta
&\implies \frac{x}{3} = \frac{1}{\cos\theta} \\
&\implies \cos\theta = \frac{3}{x} \\
&\implies \cos^2\theta = \frac{9}{x^2} \\
&\implies \sin^2\theta = 1 - \cos^2\theta = 1 - \frac{9}{x^2} = \frac{x^2 - 9}{x^2} \\
&\implies \sin\theta = \frac{\sqrt{x^2 - 9}}{x}.
\end{aligned} x = 3 sec θ ⟹ 3 x = cos θ 1 ⟹ cos θ = x 3 ⟹ cos 2 θ = x 2 9 ⟹ sin 2 θ = 1 − cos 2 θ = 1 − x 2 9 = x 2 x 2 − 9 ⟹ sin θ = x x 2 − 9 . So the integral is
1 9 sin θ + C = x 2 − 9 9 x + C . \frac{1}{9} \sin\theta + C
= \boxed{\frac{\sqrt{x^2 - 9}}{9x} + C}. 9 1 sin θ + C = 9 x x 2 − 9 + C .
Example 4.
Calculate ∫ x 2 + 2 x d x \displaystyle \int \sqrt{x^2 + 2x} \,\diff{x} ∫ x 2 + 2 x d x
Solution.
If we want to use a trig substitution, we need to get it in the form u 2 + a 2 u^2 + a^2 u 2 + a 2 u u u x x x a a a complete the square :
x 2 + 2 x = ( x + 1 ) 2 − 1. x^2 + 2x = \p{x + 1}^2 - 1. x 2 + 2 x = ( x + 1 ) 2 − 1. So, we can use x + 1 = sec θ x + 1 = \sec\theta x + 1 = sec θ d x = sec θ tan θ d θ \diff{x} = \sec\theta\tan\theta \,\diff{\theta} d x = sec θ tan θ d θ ( x + 1 ) 2 − 1 = sec 2 θ − 1 = tan 2 θ \p{x + 1}^2 - 1 = \sec^2\theta - 1 = \tan^2\theta ( x + 1 ) 2 − 1 = sec 2 θ − 1 = tan 2 θ
∫ x 2 + 2 x d x = ∫ ( x + 1 ) 2 − 1 d x = ∫ tan 2 θ sec θ tan θ d θ = ∫ sec θ tan 2 θ d θ . \begin{aligned}
\int \sqrt{x^2 + 2x} \,\diff{x}
= \int \sqrt{\p{x + 1}^2 - 1} \,\diff{x}
&= \int \sqrt{\tan^2\theta} \sec\theta\tan\theta \,\diff\theta \\
&= \int \sec\theta\tan^2\theta \,\diff\theta.
\end{aligned} ∫ x 2 + 2 x d x = ∫ ( x + 1 ) 2 − 1 d x = ∫ tan 2 θ sec θ tan θ d θ = ∫ sec θ tan 2 θ d θ . From here, we're going to end up using the same trick as calculating something like ∫ e x sin x d x \int e^x \sin{x} \,\diff{x} ∫ e x sin x d x
u = tan θ ⟹ d u = sec 2 θ d θ d v = sec θ tan θ d θ ⟹ v = sec θ , \begin{aligned}
u = \tan\theta
&\implies \diff{u} = \sec^2\theta \,\diff\theta \\
\diff{v} = \sec\theta\tan\theta \,\diff\theta
&\implies v = \sec\theta,
\end{aligned} u = tan θ d v = sec θ tan θ d θ ⟹ d u = sec 2 θ d θ ⟹ v = sec θ , which turns the integral into
∫ sec θ tan 2 θ d θ = sec θ tan θ − ∫ sec 3 θ d θ = sec θ tan θ − ∫ sec θ ( tan 2 θ + 1 ) d θ = sec θ tan θ − ∫ sec θ d θ − ∫ sec θ tan 2 θ d θ ⟹ 2 ∫ sec θ tan 2 θ d θ = sec θ tan θ − ln ∣ sec θ + tan θ ∣ + C ⟹ ∫ sec θ tan 2 θ d θ = 1 2 sec θ tan θ − 1 2 ln ∣ sec θ + tan θ ∣ + C . \begin{aligned}
\int \sec\theta\tan^2\theta \,\diff\theta
&= \sec\theta\tan\theta - \int \sec^3\theta \,\diff\theta \\
&= \sec\theta\tan\theta - \int \sec\theta\p{\tan^2\theta + 1} \,\diff\theta \\
&= \sec\theta\tan\theta - \int \sec\theta \,\diff\theta - \int \sec\theta\tan^2\theta \,\diff\theta \\
\implies
2\int \sec\theta\tan^2\theta \,\diff\theta
&= \sec\theta\tan\theta - \ln\abs{\sec\theta + \tan\theta} + C \\
\implies
\int \sec\theta\tan^2\theta \,\diff\theta
&= \frac{1}{2}\sec\theta\tan\theta - \frac{1}{2}\ln\abs{\sec\theta + \tan\theta} + C.
\end{aligned} ∫ sec θ tan 2 θ d θ ⟹ 2 ∫ sec θ tan 2 θ d θ ⟹ ∫ sec θ tan 2 θ d θ = sec θ tan θ − ∫ sec 3 θ d θ = sec θ tan θ − ∫ sec θ ( tan 2 θ + 1 ) d θ = sec θ tan θ − ∫ sec θ d θ − ∫ sec θ tan 2 θ d θ = sec θ tan θ − ln ∣ sec θ + tan θ ∣ + C = 2 1 sec θ tan θ − 2 1 ln ∣ sec θ + tan θ ∣ + C . Finally, we need to undo the substitution. We know that x + 1 = sec θ x + 1 = \sec\theta x + 1 = sec θ
( x + 1 ) 2 − 1 = tan 2 θ ⟹ tan θ = ( x + 1 ) 2 − 1 = x 2 + 2 x . \p{x + 1}^2 - 1 = \tan^2\theta
\implies \tan\theta = \sqrt{\p{x + 1}^2 - 1} = \sqrt{x^2 + 2x}. ( x + 1 ) 2 − 1 = tan 2 θ ⟹ tan θ = ( x + 1 ) 2 − 1 = x 2 + 2 x . Substituting these in, we get
∫ x 2 + 2 x d x = 1 2 sec θ tan θ − 1 2 ln ∣ sec θ + tan θ ∣ + C = 1 2 ( x + 1 ) x 2 + 2 x − 1 2 ln ∣ x + 1 + x 2 + 2 x ∣ + C . \begin{aligned}
\int \sqrt{x^2 + 2x} \,\diff{x}
&= \frac{1}{2}\sec\theta\tan\theta - \frac{1}{2}\ln\abs{\sec\theta + \tan\theta} + C \\
&= \boxed{\frac{1}{2}\p{x + 1}\sqrt{x^2 + 2x} - \frac{1}{2}\ln\abs{x + 1 + \sqrt{x^2 + 2x}} + C}.
\end{aligned} ∫ x 2 + 2 x d x = 2 1 sec θ tan θ − 2 1 ln ∣ sec θ + tan θ ∣ + C = 2 1 ( x + 1 ) x 2 + 2 x − 2 1 ln ∣ ∣ x + 1 + x 2 + 2 x ∣ ∣ + C .
Example 5.
Calculate ∫ x 2 + 1 ( x 2 − 2 x + 2 ) 2 \displaystyle \int \frac{x^2 + 1}{\p{x^2 - 2x + 2}^2} ∫ ( x 2 − 2 x + 2 ) 2 x 2 + 1
Solution.
Like before, we want to complete the square. x 2 − 2 x + 2 = ( x − 1 ) 2 + 1 x^2 - 2x + 2 = \p{x - 1}^2 + 1 x 2 − 2 x + 2 = ( x − 1 ) 2 + 1 x − 1 = tan θ x - 1 = \tan\theta x − 1 = tan θ
d x = sec 2 θ d θ x 2 − 2 x + 2 = ( x − 1 ) 2 + 1 = sec 2 θ x 2 + 1 = ( tan θ + 1 ) 2 + 1 = tan 2 θ + 2 tan θ + 2. \begin{aligned}
\diff{x}
&= \sec^2\theta \,\diff\theta \\
x^2 - 2x + 2
&= \p{x - 1}^2 + 1 = \sec^2\theta \\
x^2 + 1
&= \p{\tan\theta + 1}^2 + 1 = \tan^2\theta + 2\tan\theta + 2.
\end{aligned} d x x 2 − 2 x + 2 x 2 + 1 = sec 2 θ d θ = ( x − 1 ) 2 + 1 = sec 2 θ = ( tan θ + 1 ) 2 + 1 = tan 2 θ + 2 tan θ + 2. Substituting into the integral,
∫ x 2 + 1 ( x 2 − 2 x + 2 ) 2 = ∫ tan 2 θ + 2 tan θ + 2 sec 4 θ sec 2 θ d θ = ∫ tan 2 θ + 2 tan θ + 2 sec 2 θ d θ = ∫ ( sin 2 θ cos 2 θ + 2 sin θ cos θ + 2 ) cos 2 θ d θ = ∫ sin 2 θ + 2 sin θ cos θ + 2 cos 2 θ d θ = ∫ 1 − cos 2 θ 2 + 2 sin θ cos θ + ( 1 + cos 2 θ ) d θ = 1 2 θ − 1 4 sin 2 θ + sin 2 θ + θ + 1 2 sin 2 θ + C = 3 2 θ + 1 4 sin 2 θ + sin 2 θ + C = 3 2 θ + 1 2 sin θ cos θ + sin 2 θ + C . \begin{aligned}
\int \frac{x^2 + 1}{\p{x^2 - 2x + 2}^2}
&= \int \frac{\tan^2\theta + 2\tan\theta + 2}{\sec^4\theta} \sec^2\theta \,\diff\theta \\
&= \int \frac{\tan^2\theta + 2\tan\theta + 2}{\sec^2\theta} \,\diff\theta \\
&= \int \p{\frac{\sin^2\theta}{\cos^2\theta} + 2\frac{\sin\theta}{\cos\theta} + 2} \cos^2\theta \,\diff\theta \\
&= \int \sin^2\theta + 2\sin\theta\cos\theta + 2\cos^2\theta \,\diff\theta \\
&= \int \frac{1 - \cos{2\theta}}{2} + 2\sin\theta\cos\theta + \p{1 + \cos{2\theta}} \,\diff\theta \\
&= \frac{1}{2}\theta - \frac{1}{4} \sin{2\theta} + \sin^2\theta + \theta + \frac{1}{2} \sin{2\theta} + C \\
&= \frac{3}{2}\theta + \frac{1}{4} \sin{2\theta} + \sin^2\theta + C \\
&= \frac{3}{2}\theta + \frac{1}{2} \sin\theta\cos\theta + \sin^2\theta + C.
\end{aligned} ∫ ( x 2 − 2 x + 2 ) 2 x 2 + 1 = ∫ sec 4 θ tan 2 θ + 2 tan θ + 2 sec 2 θ d θ = ∫ sec 2 θ tan 2 θ + 2 tan θ + 2 d θ = ∫ ( cos 2 θ sin 2 θ + 2 cos θ sin θ + 2 ) cos 2 θ d θ = ∫ sin 2 θ + 2 sin θ cos θ + 2 cos 2 θ d θ = ∫ 2 1 − cos 2 θ + 2 sin θ cos θ + ( 1 + cos 2 θ ) d θ = 2 1 θ − 4 1 sin 2 θ + sin 2 θ + θ + 2 1 sin 2 θ + C = 2 3 θ + 4 1 sin 2 θ + sin 2 θ + C = 2 3 θ + 2 1 sin θ cos θ + sin 2 θ + C . From our substitution, x − 1 = tan θ ⟹ θ = arctan ( x − 1 ) x - 1 = \tan\theta \implies \theta = \arctan\p{x - 1} x − 1 = tan θ ⟹ θ = arctan ( x − 1 )
x 2 − 2 x + 2 = sec 2 θ ⟹ cos 2 θ = 1 x 2 − 2 x + 2 ⟹ cos θ = 1 x 2 − 2 x + 2 , \begin{aligned}
x^2 - 2x + 2 = \sec^2\theta
&\implies \cos^2\theta = \frac{1}{x^2 - 2x + 2} \\
&\implies \cos\theta = \frac{1}{\sqrt{x^2 - 2x + 2}},
\end{aligned} x 2 − 2 x + 2 = sec 2 θ ⟹ cos 2 θ = x 2 − 2 x + 2 1 ⟹ cos θ = x 2 − 2 x + 2 1 , and
sin 2 θ = 1 − cos 2 θ = 1 − 1 x 2 − 2 x + 2 = x 2 − 2 x + 1 x 2 − 2 x + 2 ⟹ sin θ = x 2 − 2 x + 1 x 2 − 2 x + 2 = ( x − 1 ) 2 x 2 − 2 x + 2 = x − 1 x 2 − 2 x + 2 \begin{aligned}
\sin^2\theta
&= 1 - \cos^2\theta = 1 - \frac{1}{x^2 - 2x + 2} = \frac{x^2 - 2x + 1}{x^2 - 2x + 2} \\
\implies
\sin\theta
&= \sqrt{\frac{x^2 - 2x + 1}{x^2 - 2x + 2}} = \sqrt{\frac{\p{x - 1}^2}{x^2 - 2x + 2}} = \frac{x - 1}{\sqrt{x^2 - 2x + 2}}
\end{aligned} sin 2 θ ⟹ sin θ = 1 − cos 2 θ = 1 − x 2 − 2 x + 2 1 = x 2 − 2 x + 2 x 2 − 2 x + 1 = x 2 − 2 x + 2 x 2 − 2 x + 1 = x 2 − 2 x + 2 ( x − 1 ) 2 = x 2 − 2 x + 2 x − 1 So, substituting all of these into our integral,
3 2 θ + 1 2 sin θ cos θ + sin 2 θ + C = 3 2 arctan ( x − 1 ) + 1 2 1 x 2 − 2 x + 2 x − 1 x 2 − 2 x + 2 + x 2 − 2 x + 1 x 2 − 2 x + 2 + C = 3 2 arctan ( x − 1 ) + x − 1 2 ( x 2 − 2 x + 2 ) + x 2 − 2 x + 1 x 2 − 2 x + 2 + C = 3 2 arctan ( x − 1 ) + x − 1 2 ( x 2 − 2 x + 2 ) + 2 x 2 − 4 x + 2 2 ( x 2 − 2 x + 2 ) + C = 3 2 arctan ( x − 1 ) + 2 x 2 − 3 x + 1 2 x 2 − 4 x + 4 + C . \begin{aligned}
&\frac{3}{2}\theta + \frac{1}{2} \sin\theta\cos\theta + \sin^2\theta + C \\
={}& \frac{3}{2} \arctan\p{x - 1} + \frac{1}{2} \frac{1}{\sqrt{x^2 - 2x + 2}}\frac{x - 1}{\sqrt{x^2 - 2x + 2}} + \frac{x^2 - 2x + 1}{x^2 - 2x + 2} + C \\
={}& \frac{3}{2} \arctan\p{x - 1} + \frac{x - 1}{2\p{x^2 - 2x + 2}} + \frac{x^2 - 2x + 1}{x^2 - 2x + 2} + C \\
={}& \frac{3}{2} \arctan\p{x - 1} + \frac{x - 1}{2\p{x^2 - 2x + 2}} + \frac{2x^2 - 4x + 2}{2\p{x^2 - 2x + 2}} + C \\
={}& \boxed{\frac{3}{2} \arctan\p{x - 1} + \frac{2x^2 - 3x + 1}{2x^2 - 4x + 4} + C}.
\end{aligned} = = = = 2 3 θ + 2 1 sin θ cos θ + sin 2 θ + C 2 3 arctan ( x − 1 ) + 2 1 x 2 − 2 x + 2 1 x 2 − 2 x + 2 x − 1 + x 2 − 2 x + 2 x 2 − 2 x + 1 + C 2 3 arctan ( x − 1 ) + 2 ( x 2 − 2 x + 2 ) x − 1 + x 2 − 2 x + 2 x 2 − 2 x + 1 + C 2 3 arctan ( x − 1 ) + 2 ( x 2 − 2 x + 2 ) x − 1 + 2 ( x 2 − 2 x + 2 ) 2 x 2 − 4 x + 2 + C 2 3 arctan ( x − 1 ) + 2 x 2 − 4 x + 4 2 x 2 − 3 x + 1 + C .
Homework
Here are some hints for the problems this week:
8.1.69
This is just a volume of solids of revolution problem. You end up needing to calculate
∫ 0 π x 2 sin x d x , \int_0^\pi x^2 \sin{x} \,\diff{x}, ∫ 0 π x 2 sin x d x , which you can do by integrating by parts twice.
8.1.97
For (a), you should end up with
I n = 1 2 x n − 1 sin x 2 − n − 1 2 J n − 2 J n = − 1 2 x n − 1 cos x 2 + n − 1 2 I n − 2 . \begin{aligned}
I_n
&= \frac{1}{2} x^{n-1} \sin{x^2} - \frac{n - 1}{2} J_{n-2} \\
J_n
&= -\frac{1}{2} x^{n-1} \cos{x^2} + \frac{n - 1}{2} I_{n-2}.
\end{aligned} I n J n = 2 1 x n − 1 sin x 2 − 2 n − 1 J n − 2 = − 2 1 x n − 1 cos x 2 + 2 n − 1 I n − 2 . The idea for (b) is this: if you can calculate I 1 I_1 I 1 J 3 J_3 J 3 J 1 J_1 J 1 I 3 I_3 I 3
I 1 → J 3 → I 5 → J 7 → ⋯ J 1 → I 3 → J 5 → I 7 → ⋯ \begin{aligned}
I_1 \to J_3 \to I_5 \to J_7 \to \cdots \\
J_1 \to I_3 \to J_5 \to I_7 \to \cdots
\end{aligned} I 1 → J 3 → I 5 → J 7 → ⋯ J 1 → I 3 → J 5 → I 7 → ⋯ So, if you can calculate I 1 I_1 I 1 J 1 J_1 J 1 I n I_n I n J n J_n J n n n n I 1 I_1 I 1 J 1 J_1 J 1
8.2.23
For this, you can try the reduction formulas, or you can just use the trig identity
cos 2 θ = 1 + cos 2 θ 2 \cos^2\theta = \frac{1 + \cos{2\theta}}{2} cos 2 θ = 2 1 + cos 2 θ twice. For example, after using it once, you get
∫ cos 4 ( 3 x + 2 ) d x = ∫ ( 1 + cos ( 6 x + 4 ) 2 ) 2 d x = 1 4 ∫ 1 + 2 cos ( 6 x + 4 ) + cos 2 ( 6 x + 4 ) d x . \begin{aligned}
\int \cos^4\p{3x + 2} \,\diff{x}
&= \int \p{\frac{1 + \cos\p{6x + 4}}{2}}^2 \,\diff{x} \\
&= \frac{1}{4} \int 1 + 2\cos\p{6x + 4} + \cos^2\p{6x + 4} \,\diff{x}.
\end{aligned} ∫ cos 4 ( 3 x + 2 ) d x = ∫ ( 2 1 + cos ( 6 x + 4 ) ) 2 d x = 4 1 ∫ 1 + 2 cos ( 6 x + 4 ) + cos 2 ( 6 x + 4 ) d x .