Tag Archives: integration by parts

Integration Example Problem 1

The following indefinite integral is a great exercise for students in introductory calculus since it requires most of the techniques they learn in order to integrate.

First rewrite the integral in order to use a trigonometric substitution.

$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \int \sqrt{ \left( x + 3 \right)^2 – 4 } \ dx $$

Substituting and simplifying yields

$$ x = 2 \sec \theta – 3 \hspace{ 20 mm } dx = 2 \sec \theta \tan \theta \ d \theta $$
$$ 2 \int \sqrt{ 4 \sec^2 \theta – 4 } \sec \theta \tan \ \theta d \theta $$
$$ 4 \left( \int \tan^2 \theta \sec \theta \ d \theta \right) = 4 \left( \int \left( \sec^2 \theta – 1 \right) \sec \theta \ d \theta \right) $$

From here we need to use integration by parts and the antiderivative of \( \sec \theta \). I assume the \( \int \sec \theta \ d \theta \) is known to the reader. If not, click here for a pop-out window solution.

$$4 \left( \int \sec^3 \theta \ d \theta – \int \sec \theta \ d \theta \right) = 4 \left( \int \sec^3 \theta \ d \theta – \ln \left| \sec \theta + \tan \theta \right| + C \right) $$

Using integration by parts

$$ \int \sec^3 \theta \ d \theta = \int \sec^2 \theta \sec \theta \ d \theta $$
$$ u = \sec \theta \hspace{20 mm } du = \sec \theta \tan \theta \ d \theta $$
$$ dv = \sec^2 \theta \ d \theta \hspace{20 mm } v = \tan \theta $$
$$ \int \sec^3 \theta \ d \theta = \sec \theta \tan \theta \ – \int \tan^2 \theta \sec \theta \ d \theta $$
$$ \int \sec^3 \theta \ d \theta = \sec \theta \tan \theta \ – \int \sec^3 \theta + \int \sec \theta \ d \theta $$
$$ \int \sec^3 \theta \ d \theta = \frac{1}2 \sec \theta \tan \theta + \frac{1}2 \ln \left| \sec \theta + \tan \theta \right| $$

Remembering to multiply through by 4

$$ 4 \int \sec^3 \theta \ d \theta = 2 \sec \theta \tan \theta + 2 \ln \left| \sec \theta + \tan \theta \right| $$

Reassembling the pieces of the original integral and simplifying

$$ 4 \left( \int \tan^2 \theta \sec \theta \ d \theta \right) = 2 \sec \theta \tan \theta + 2 \ln \left| \sec \theta + \tan \theta \right| \ – 4 \ln \left| \sec \theta + \tan \theta \right| + C $$
$$ 4 \left( \int \tan^2 \theta \sec \theta \ d \theta \right) = 2 \sec \theta \tan \theta – 2 \ln \left| \sec \theta + \tan \theta \right| + C $$

Substituting back in for \( x \)

$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \frac{ x + 3}{2} \sqrt{ x^2 + 6x + 5 } – 2 \ln \left| \frac{ x + 3 }{2} + \frac{ \sqrt{ x^2 + 6x + 5 } }2 \right| + K $$

Using log rules the ratio can be removed from the argument of the log function.

$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \frac{ x + 3}{2} \sqrt{ x^2 + 6x + 5 } – 2 \left( \ln \left| x + 3 + \sqrt{ x^2 + 6x + 5 } \right| – \ln \left| 2 \right| \right) + K $$
$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \frac{ x + 3}{2} \sqrt{ x^2 + 6x + 5 } – 2 \ln \left| x + 3 + \sqrt{ x^2 + 6x + 5 } \right| + 2 \ln \left| 2 \right| + K $$
$$ \int \sqrt{ x^2 + 6x + 5 } \ dx = \frac{ x + 3}{2} \sqrt{ x^2 + 6x + 5 } – 2 \ln \left| x + 3 + \sqrt{ x^2 + 6x + 5 } \right| + C_1$$

where \( C_1 = 2 \ln \left| 2 \right| + K \)

Reduction Formulas

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\( \int \! \sin^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 3 \)

\( \int \! \sin^n x \ dx = \int \! \left( \sin^{n-1} x \right) \sin x \ dx \)

\( u = \sin^{n-1} x \)

\( du = \left(n-1 \right) \sin^{n-2} x \cos x \ dx \)

\( dv = \sin x \ dx \)

\( v = – \cos x \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \cos^2 x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \left( 1 – \sin^2 x \right) \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x – \sin^{\left(n-2\right)+2} x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x – \sin^n x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \ dx – \left(n-1 \right) \int \! \sin^n x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx + \left(n-1 \right) \int \! \sin^n x \ dx= -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \ dx \)

\( n \int \! \left( \sin^{n-1} x \right) \sin x \ dx = -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \ dx \)

\( \int \! \left( \sin^{n-1} x \right) \sin x \ dx = \frac{1}n \left( -\cos x \sin^{n-1} x + \left( n-1 \right) \int \! \sin^{n-2} x \ dx \right) \)

 

\( \int \! \cos^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 3 \)

\( \int \! \cos^n x \ dx = \int \! \left( \cos^{n-1} x \right) \cos x \ dx \)

\( u = \cos^{n-1} x \)

\( du = – \left(n-1 \right) \cos^{n-2} x \sin x \ dx \)

\( dv = \cos x \ dx \)

\( v = \sin x \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \sin^2 x \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \left( 1 – \cos^2 x \right) \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x – \cos^{\left(n-2\right)+2} x \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x – \cos^n x \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \ dx – \left(n-1 \right) \int \! \cos^n x \ dx \)

\(n \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \ dx \)

\( \int \! \left( \cos^{n-1} x \right) \cos x \ dx = \frac{1}n \left( \sin x \cos^{n-1} x + \left( n-1 \right) \int \! \cos^{n-2} x \ dx \right) \)

 

\( \int \! \tan^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 2 \)

\( \int \! \tan^n x \ dx = \int \! \tan^{n-2}x \tan^2 x \ dx\)

\( \int \! \tan^n x \ dx = \int \! \tan^{n-2}x \left( \sec^2 x – 1 \right) dx\)

\( \int \! \tan^n x \ dx = \int \! \tan^{n-2}x \sec^2 x \ dx – \int \! \tan^{n-2}x \ dx \)

\( \int \! \tan^{n-2}x \sec^2 x \ dx\)

\( u = \tan x \)

\( du = \sec^2 x \ dx\)

\( \int \! u^{n-2} \ du = \frac{u^{n-1}}{n-1} = \frac{ \tan^{n-1}x}{n-1} \)

\( \int \! \tan^n x \ dx = \frac{ \tan^{n-1}x}{n-1} – \int \! \tan^{n-2}x \ dx \)

 

\( \int \! \sec^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 3 \)

\( \int \! \sec^n x \ dx = \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx \)

\( u = \sec^{n-2} x \)

\( du = \left(n-2 \right) \sec^{n-3} x \sec x \tan x \ dx \)

\( dv = \sec^2 x \ dx \)

\( v = \tan x \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x – \left( n-2 \right) \int \! \tan^2 x \sec^{n-2} x \ dx \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x – \left( n-2 \right) \int \! \left( \sec^2 x – 1 \right) \sec^{n-2} x \ dx \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x – \left( n-2 \right) \int \! \sec^n x – \sec^{n-2} x \ dx \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x – \left( n-2 \right) \int \! \sec^n x \ dx + \left(n-2 \right) \int \! \sec^{n-2} x \ dx \)

\( \left(n-1 \right) \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \tan x \sec^{n-2} x + \left( n-2 \right) \int \! \sec^{n-2} x \ dx \)

\( \int \! \sec^2 x \left( \sec^{n-2} x \right) \ dx = \frac{1}{n-1} \left( \tan x \sec^{n-2} x + \left( n-2 \right) \int \! \sec^{n-2} x \ dx \right) \)

 

\( \int \! \csc^n x \ dx \)       where \( n \in \mathbb{N} \) and \( n \ge 3 \)

\( \int \! \csc^n x \ dx = \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx \)

\( u = \csc^{n-2} x \)

\( du = \left(n-2 \right) \csc^{n-3} x \left(-\csc x \cot x \right) \ dx \)

\( dv = \csc^2 x \ dx \)

\( v = -\cot x \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x – \left( n-2 \right) \int \! \cot^2 x \csc^{n-2} x \ dx \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x – \left( n-2 \right) \int \! \left( \csc^2 x – 1 \right) \sec^{n-2} x \ dx \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x – \left( n-2 \right) \int \! \csc^n x – \csc^{n-2} x \ dx \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x – \left( n-2 \right) \int \! \csc^n x \ dx + \left(n-2 \right) \int \! \csc^{n-2} x \ dx \)

\( \left(n-1 \right) \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = -\cot x \csc^{n-2} x + \left( n-2 \right) \int \! \csc^{n-2} x \ dx \)

\( \int \! \csc^2 x \left( \csc^{n-2} x \right) \ dx = \frac{1}{n-1} \left( -\cot x \csc^{n-2} x + \left( n-2 \right) \int \! \csc^{n-2} x \ dx \right) \)

Integration by Parts Problems

$$ \int \! u \ dv = uv \ – \int \! v \ du$$

\( \int \! \ln x \ dx \)

\( u = \ln x \ \ \ \ \ \ du = \frac{1}{x} \ dx \)

\( dv = dx \ \ \ \ \ \ v = x \)

\( \int \! \ln x \ dx = x \ln x – \int \! x \frac{1}{x} \ dx \)

\( \int \! \ln x \ dx = x \ln x – x + C \)

 

\( \int \! x \ln x \ dx \)

\( u = \ln x \ \ \ \ \ \ du = \frac{1}{x} \ dx \)

\( dv = x \ dx \ \ \ \ \ \ v = \frac{x^2}{2}\)

\( \int \! x \ln x \ dx = \frac{x^2}{2} \ln x – \int \! \frac{x^2}{2} \frac{1}{x} \ dx \)

\( \int \! x \ln x \ dx = \frac{x^2}{2} \ln x – \frac{x^2}{4} + C \)

 

\( \int \! \sin^{-1} x \ dx \)

\( u = \sin^{-1} x \ \ \ \ \ \ du = \frac{1}{\sqrt{1 – x^2}} \ dx\)

\( dv = dx \ \ \ \ \ \ v = x \)

\( \int \! \sin^{-1} x \ dx = x \sin^{-1} x – \int \! x \frac{1}{ \sqrt{1 – x^2 }} \ dx\)

\( \int \! \frac{x}{ \sqrt{1 – x^2 }} \ dx\)

\( t = 1 – x^2 \ \ \ \ \ \ – \frac{1}{2} \ dt = x \ dx\)

\( – \int \! \frac{1}{ 2 \sqrt{ t }} \ dt = – \sqrt{ t } + C\)

\( \int \! \sin^{-1} x \ dx = x \sin^{-1} x – \int \! x \frac{1}{ \sqrt{1 – x^2 }} \ dx = x \sin^{-1} x + \sqrt{ 1 – x^2 } + C \)

 

\( \int \! x e^{ x } \ dx \)

\( u = x \ \ \ \ \ \ du = dx \)

\( dv = e^{x} \ dx \ \ \ \ \ \ v = e^{x}\)

\( \int \! x e^{ x } \ dx = x e^{x} – \int \! e^{x} \ dx = x e^{x} – e^{x} + C\)

 

\( \int \! e^{ \sqrt{x} } \ dx \)

\( t = \sqrt{x} \ \ \ \ \ \ dt = \frac{1}{2 \sqrt{ x }} \ dx\)

\( 2 t \ dt = dx \)

\( \int \! 2 t e^{ t } \ dt \)

\( u = 2 t \ \ \ \ \ \ du = 2 \ dt \)

\( dv = e^{t} \ dt \ \ \ \ \ \ v = e^{t}\)

\( \int \! 2 t e^{ t } \ dt = 2 t e^{t} – \int \! 2 e^{t} \ dt = 2 t e^{t} – 2 e^{t} + K\)

\( \int \! e^{ \sqrt{x} } \ dx = 2 \sqrt{ x } e^{ \sqrt{ x } } -2 e^{ \sqrt{ x } } + C\)

 

\( \int \! e^{ x } \sin x \ dx \)

\( u = \sin x \ \ \ \ \ \ dv = e^{ x } \ dx\)

\( du = \cos x \ dx \ \ \ \ \ \ v = e^{ x } \)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – \int \! e^{ x } \cos x \ dx \)

\( s = \cos x \ \ \ \ \ \ ds = – \sin x \ dx \)

\( dt = e^{x} \ dx \ \ \ \ \ \ t = e^{x}\)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – \left( e^{ x } \cos x – \int \! – e^{ x } \sin x \ dx \right)\)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – e^{ x } \cos x – \int \! e^{ x } \sin x \ dx \)

\( \int \! e^{ x } \sin x \ dx = \frac{1}{2} \left( e^{ x } \sin x – e^{ x } \cos x \right) + C \)