Calculate integrals using different techniques with examples and detailed solutions and explanations. Also more exercises with solutions are presented at the bottom of the page.
In all examples and exercises, c represent the constant of integration.
Example 1
Evaluate the integral
∫ 6 cos x sin x dx
∫ 6 cos x sin x dx= - (3/2) cos (2 x) + c
As an exercise, differentiate - (3/2) cos 2 x + c to obtain 6 sin x cos x which is the integrand in the given integral. This is a way to check the answer to integrals calculations. \( \) \( \)\( \)\( \)\( \)
Example 2
Calculate the integral
\[ \displaystyle\int x \; \sqrt{x+1} dx \]
Solution to Example 2:
Use Integration by Substitution: Let \( u = x + 1 \) which gives \( x = u - 1 \).
\( \displaystyle\int x \sqrt{x+1} \; dx = \int (u-1) \cdot u^{1/2} \; dx \)
The above leads to \( \dfrac{du}{dx}=1 \) and \( du = dx \) and the given integral is equal to
\( = \displaystyle\int (u^{3/2}-u^{1/2}) \; du \)
We now use the property for the integral of the sum of functions and the formula for integration of power function: \( \displaystyle \int x^n dx = \left(\dfrac{1}{n+1}\right) \; x^{n+1} + c \) to obtain
\( \displaystyle = (2 / 5) u^{5/2} - (2 / 3) u^{3/2} + c \)
We now substitute \( u \) by \( x + 1 \) into the above result to obtain the final result as follows
\[ \displaystyle\int x \sqrt{x+1} dx = (2 / 5) (x + 1)^{5/2} - (2 / 3) (x + 1)^{3/2} + c \]
To check the final answer, differentiate the indefinite integral obtained to obtain the integrand \( x \sqrt{x + 1} \) in the given integral.
Example 3
Evaluate the integral
\[ \displaystyle \int \cos^2 x \; dx \]
Solution to Example 3
Use the trigonometric identity \( \cos^2 x = \dfrac{1+\cos(2x)}{2} \) to rewrite the given integral as
\( \displaystyle\int \cos^2 x \; dx = \int \dfrac{1+cos(2x)}{2} \; dx \)
Use Integration by Substitution: \( u = 2x \) so that \( du = 2 dx \) and \( dx = du / 2 \), and the given integral can be written as
\( = \displaystyle \int (1/4)(1+\cos u) \; du \)
Integrate to obtain
\( = (1 / 4) u + (1 / 4) \sin (u) + c \)
Substitute back \( u \) by \( 2 x \) and simplify
\[ \displaystyle \int \cos^2 x \; dx = x / 2 + (1 / 4) \sin (2 x) + c \]
Check the final answer by differentiation.
Example 4
Evaluate the integral
\[\displaystyle\int x^3 e^{x^4} \; dx\]
Solution to Example 4
Use Integration by Substitution: Let \( u = x^4\) so that \( du / dx = 4 x^3 \) which leads to \( (1 / 4) du = x^3 dx \), so that the given integral can be written as
\( \displaystyle\int x^3 e^{x^4} dx = \int (1 / 4) e^u \; du \)
We now use the formula for the integral of the exponential function to write
\( \displaystyle\int (1 / 4) e^u \; du = (1 / 4) e^u + c \)
Substitute back \( u \) by \( x^4 \) to obtain the final answer
\[ \displaystyle\int x^3 e^{x^4} dx = (1 / 4) e^{x^4}+ c \]
Example 5
Calculate the integral
\[\displaystyle \int \dfrac{\sin (2x)}{1-\cos^2(x)} \; dx \]
Solution to Example 5
Use the trigonometric identities: \( \sin (2x) = 2 \sin x \; \cos x \) and \( 1-\cos^2(x) = \sin^2(x) \) to rewrite the given integral as
\( \displaystyle \int \dfrac{\sin (2x)}{1-\cos^2(x)}dx = \displaystyle \int \dfrac{2 \sin x \cos x}{\sin^2(x)}dx \)
Simplify and rewrite as
\( = \displaystyle \int 2 \dfrac{\cos x}{\sin x} \; dx \)
Use the formula of integration \( \displaystyle\int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \) to obtain
\( = 2 \ln |\sin \; x| \)
Use the log property \( \; a \ln x = \ln x^a \) to rewrite the final result as
\[ \displaystyle \int \dfrac{\sin (2x)}{1-\cos^2(x)}dx = \ln \sin^2 \; x \]
Example 6
Calculate the integral
\[\displaystyle \int (x+\sin x)^2 \; dx \]
Solution to Example 6
Expand \( \displaystyle (x+\sin x)^2 = x^2 + \sin^2 x + 2 x \sin x \) and apply the sum rule of integrals\( \displaystyle \int (f(x) + g(x) + h(x) ) dx = \int f(x) dx + \int g(x) dx + \int h(x) dx \) to rewrite the given integral as
\( \displaystyle \int (x+\sin x)^2 \; dx = \int x^2 \; dx + \int \sin^2 x \; dx + \int 2 x \; \sin x \; dx\)
Calculate each integral in the sum above
\( \displaystyle \int x^2 \; dx = (1/3) x^3 + c\)
Use the trigonometric identity \( \; \sin^2 x = (1 – \cos (2x)) / 2 \) to reduce the power and rewrite the second integral in the sum as
\( \displaystyle \int \sin^2 x \; dx = \int \dfrac{1 – \cos (2x)}{2} \; dx = (1/2) x - (1/4) \sin (2x) + c\)
The third integral is calculated using the integration by parts: \( \int u' v \; dx = u v - \int u v' \; dx \).
Let \( v = x \) and \( u' = \sin x \) so that \( v' = 1 \) simplifies calculations and \( u = - \cos x\).
\( \displaystyle \int 2 x \; \sin x \; dx = 2 (- x \cos x - \int (- cos x) dx ) = - 2 x \cos x + 2 \sin x + c \)
The final answer to the given integral is given by the sum of all three integrals calculated above
\[\displaystyle \int (x+\sin x)^2 \; dx = (1/3) x^3 + (1/2) x - (1/4) \sin (2x) - 2 x \cos x + 2 \sin x + c \]
Example 7
Calculate the integral
\[\displaystyle \int \dfrac{\sin x}{\sin^2 x - 2 cos x + 2} \; dx \]
Solution to Example 7
Use the trigonometric identity \( \sin^2 x = 1 - \cos^2 x \) to rewrite the given integral as
\(\displaystyle \int \dfrac{\sin x}{\sin^2 x - 2 \cos x + 2} \; dx = \int \dfrac{\sin x}{1 - \cos^2 x - 2 \cos x + 2} \; dx \)
Use Integration by Substitution: Let \( u = \cos x \) which gives \( \dfrac{du}{dx} = - \sin x \) and \( du = - \sin x \; dx \) and substitute in the integral
\( \displaystyle \int \dfrac{\sin x}{1 - \cos^2 x - 2 \cos x + 2} \; dx = \int \dfrac{1}{u^2 + 2 u - 3} du\)
Factor the denominator \( u^2 + 2 u - 3 = (u+3)(u-1) \) and use partial fractions decomposition to rewrite \( \dfrac{1}{u^2 + 2 u - 3} \) as
\( \dfrac{1}{u^2 + 2 u - 3} = \dfrac{A}{u+3} + \dfrac{B}{u+3} \\\\
= \dfrac{1}{4(u-1)}-\dfrac{1}{4(u+3)}\)
Substitute in the integral in \( u \) and use the sum rule of integrals to write
\( \displaystyle \int \dfrac{1}{u^2 + 2 u - 3} du = \int \dfrac{1}{4(u-1)} du - \int \dfrac{1}{4(u+3)} du \)
Use formula of integration \( \displaystyle\int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \) to each integral
\( = (1/4) (\ln |u-1| - \ln|u+3| + c \)
Substitute \( u = \cos x \) to obtain the final resul as
\[\displaystyle \int \dfrac{\sin x}{\sin^2 x - 2 cos x + 2} \; dx = (1/4) (\ln | \cos x -1| - \ln| \cos x +3| ) + c \]
Usin the log properties \( \ln \dfrac {X}{Y} = \ln X - \ln Y \) to rewrite the final answer as
\[\displaystyle \int \dfrac{\sin x}{\sin^2 x - 2 cos x + 2} \; dx = (1/4) (\ln |\dfrac{ \cos x -1}{\cos x +3} | + c \]
Example 8
Calculate the integral
\[\displaystyle \int \dfrac{1}{x+\sqrt{x+2}} \; dx \]
Solution to Example 8
Use Integration by Substitution: Let \( u = \sqrt {x + 2} \); square both sides and solve for \( x \) to obtain \( x = u^2 - 2 \).
We also have \( \dfrac{du}{dx} = \dfrac{1}{2} (x+2)^{-1/2} \) which gives \( dx = 2 (x+2)^{ 1/2} \; du = 2u \; du\) . The integral becomes
\( \displaystyle \int \dfrac{1}{x+\sqrt{x+2}} \; dx = 2 \int \dfrac {u}{u^2 -2 + u} \; du \)
Factor the denominator \( u^2 -2 + u = (u+2)(u-1) \) and write the partial fraction of \( \dfrac {u}{u^2 -2 + u} \) as
\( \dfrac {u}{u^2 -2 + u} = \dfrac{A}{u-1} + \dfrac{B}{u+2} \)
Solve for \( A \) and \( B \) to obtain
\( \dfrac {u}{u^2 -2 + u} = \dfrac{1}{3(u-1)} + \dfrac{2}{3(u+2)} \)
Substitute in the integral in \( u \) and use the sum rule of integrals to write
\( \displaystyle 2 \int \dfrac {u}{u^2 -2 + u} \; du = 2 \int \dfrac{1}{3(u-1)} du + 2 \int \dfrac{2}{3(u+2)} du \)
Use formula of integration \( \displaystyle\int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \) to each integral
\( = (2/3) \ln |u-1| + (4/3) \ln|u+2| + c \)
Substitute \( u = \sqrt {x + 2} \) to obtain the final resul as
\[\displaystyle \int \dfrac{\sin x}{\sin^2 x - 2 cos x + 2} \; dx = (2/3) \ln |\sqrt {x + 2}-1| + (4/3) \ln|\sqrt {x + 2}+2| + c \]
Example 9
Evaluate the integral
\[\displaystyle \int \dfrac{1}{\tan x} \; dx \]
Solution to Example 9
Use the trigonometric identity \( \; \tan(x) = \dfrac{\sin(x)}{\cos(x)} \) to rewrite the integral as
\(\displaystyle \int \dfrac{1}{\tan x} \; dx = \int \dfrac {\cos x}{\sin x} \; dx \)
Use formula of integration \( \displaystyle\int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \) to obtain
\( = \ln | \sin x | + c \)
Hence
\[\displaystyle \int \dfrac{1}{\tan x} \; dx = \ln | \sin x | + c \]
Example 10
Evaluate the integral
\[\displaystyle \int \dfrac{1}{x^2 + 2x + 1} \; dx \]
Solution to Example 10
Complete the square in the denominator
\( \displaystyle \int \dfrac{1}{x^2 + 2x + 1} \; dx = \int \dfrac {1}{(x+1)^2} \; dx \)
Use Integration by Substitution: Let \( u = x+ 1 \) and hence \( \dfrac{du}{dx} = 1 \) which gives \( dx = du \)
\( = \displaystyle \int \dfrac {1}{u^2} \; du \)
Rewrite as
\( = \displaystyle \int u^{-2} \; du \)
Use the formula for integration of power function: \( \displaystyle \int x^n dx = \left(\dfrac{1}{n+1}\right) \; x^{n+1} + c \) to obtain
\( = - u^{-1} + c \)
\( = - \dfrac{1}{u}+ c\)
Substitute back: \( u = x+ 1 \) to write the final result as
\[ \displaystyle \int \dfrac{1}{x^2 + 2x + 1} \; dx = - \dfrac{1}{x+1} + c \]
Example 11
Evaluate the integral
\[\displaystyle \int \dfrac{1}{x^2+x+1} \; dx \]
Solution to Example 11
Note that the denominator \( x^2+x+1 \) cannot be factored over the rationals and therefore the partial fractions decomposition method cannot be used.
A known integral close to the given integral is \( \displaystyle \int \dfrac{1}{x^2+1}dx = \arctan (x) + c \).
Start by completing the square in the denominator
\( x^2+x+1 = (x + 1/2)^2 + 3/4 \)
Use Integration by Substitution: Let \( u = x + 1/2 \) which gives \( dx = du \) and rewrite the integral as
\(\displaystyle \int \dfrac{1}{x^2+x+1}dx = \int \dfrac{1}{u^2+3/4} du \)
Factor \( 3/4 \) in the denominator
\(\displaystyle = \dfrac{4}{3} \int \dfrac{1}{\dfrac{4}{3} u^2+1} du \)
Substitute: \( w = \dfrac{2}{\sqrt 3} u \) which gives \( w^2 = \dfrac{4}{ 3} u^2 \) and \( \dfrac{dw}{du} = \dfrac{2}{\sqrt 3} \) which may be written as \( du = \dfrac{\sqrt 3}{2} dw \) and rewrite the integral as
\( \displaystyle \dfrac{1}{3/4} \int \dfrac{1}{\dfrac{4}{3} u^2+1} du = \dfrac{1}{3/4} \dfrac{\sqrt 3}{2} \int \dfrac{1}{w^2+1} dw \)
Simplify and rewrite the result as
\( \displaystyle \dfrac{1}{3/4} \int \dfrac{1}{\dfrac{4}{3} u^2+1} du = \dfrac{2}{\sqrt{3}} \arctan w\)
Substitute \( w = \dfrac{2}{\sqrt 3} u \) and \( u = x + 1/2 \) or directly \( w = \dfrac{2}{\sqrt 3} (x+1/2) \) to obtain the final answer as
\[\displaystyle \int \dfrac{1}{x^2+x+1}dx = \dfrac{2}{\sqrt{3}} \arctan \left(\dfrac{2}{\sqrt 3} (x+1/2) \right) \]
Example 12
Evaluate the integral
\[\displaystyle \int \dfrac{x^4-2x^2+x}{x^2+x+1} \; dx \]
Solution to Example 12
Note that the degree of the numerator is greater than the degree of the denominator and we therefore divide the numerator by the denominator.
\( \dfrac{x^4-2x^2+x}{x^2+x+1} = x^2-x-2+\dfrac{4x+2}{x^2+x+1} \)
The given integral may be written as
\(\displaystyle\int \frac{x^4-2x^2+x}{x^2+x+1}dx = \int \left(x^2-x-2+\dfrac{4x+2}{x^2+x+1} \right) du \)
\( x^2-x-2 \) is a polynomial.
\( \dfrac{4x+2}{x^2+x+1} \) may be written as
\( \dfrac{4x+2}{x^2+x+1} = 2 \dfrac{2x+1}{x^2+x+1} \)
Note that the numerator \( 2x+1 \) is the derivative of the denominator \( x^2+x+1 \) and therefore we use the formula of integration \( \displaystyle\int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \) to obtain the final answer
\[ \displaystyle \int \frac{x^4-2x^2+x}{x^2+x+1}dx = (1/3)x^3 - (1/2)x^2 - 2 x + 2 \ln |x^2+x+1| + c \]
Example 13
Evaluate the integral
\[\displaystyle \int \left(x^3-\dfrac{1}{x^2}\right)^4 \; dx \]
Solution to Example 13
Use the binomial theorem \( \displaystyle \left(x+y\right)^n=\sum _{i=0}^n\binom{n}{i}x^{\left(n-i\right)}y^i \) to expand the integrand.
\( \left(x^3-\dfrac{1}{x^2}\right)^4 = x^{12}-4x^7+6x^2-\dfrac{4}{x^3}+\dfrac{1}{x^8} \)
The given integral may be written as
\(\displaystyle \int \left(x^3-\dfrac{1}{x^2}\right)^4 \; dx = \int \left(x^{12}-4x^7+6x^2-\dfrac{4}{x^3}+\dfrac{1}{x^8} \right) dx \)
Use integration of power function: \( \displaystyle \int x^n dx = \left(\dfrac{1}{n+1}\right) \; x^{n+1} + c \) to obtain obtain the final answer
\[ \displaystyle \left(x^3-\dfrac{1}{x^2}\right)^4 \; dx = \dfrac{x^{13}}{13}-\dfrac{x^8}{2}+2x^3+\dfrac{2}{x^2}-\dfrac{1}{7x^7}+ c \]
Example 14
Evaluate the integral
\[\displaystyle \int \tan^2(x) \; dx \]
Solution to Example 14
Use the trigonometric identity \( \tan^2 x = \sec^2 x - 1 \) to rewrite the integral as.
\(\displaystyle \int \tan^2(x) \; dx = \int (\sec^2 x - 1) \; dx \\
= \displaystyle \int \sec^2 x \; dx - \int \; dx \)
Use the formula \( \displaystyle \int \sec^2 x dx = \tan (x) + c \) to obtain the final answer
\[\displaystyle \int \tan^2(x) \; dx = \tan (x) - x + c \]
Example 15
Evaluate the integral
\[\displaystyle \int x^4(4x^5 - 2)^{10} \; dx \]
Solution to Example 15
Note that the derivative of \( 4x^5 - 2 \) is equal to \( 20 x^4\) hence the method Substitution: Let \( u = 4x^5 - 2 \) which gives \( \dfrac{du}{dx} = 20 x^4 \) and \( dx = \dfrac{du}{20 x^4} \)
The integral may be written as
\(\displaystyle \int x^4(4x^5 - 2)^{10} \; dx = \int x^4 u^{10} \dfrac{1}{20 x^4} du \)
Simplify and write as
\(\displaystyle \int x^4(4x^5 - 2)^{10} \; dx = \dfrac{1}{20} \int u^{10} du \)
Use integration of power function: \( \displaystyle \int x^n dx = \left(\dfrac{1}{n+1}\right) \; x^{n+1} + c \) to obtain
\(\displaystyle \int x^4(4x^5 - 2)^{10} \; dx = \dfrac{1}{120} u^{11} + c \)
Substitute \( u = 4x^5 - 2 \)
\[\displaystyle \int x^4(4x^5 - 2)^{10} \; dx = \dfrac{1}{220} ( 4x^5 - 2 )^{11} + c \]
Example 16
Evaluate the integral
\[\displaystyle \int x^2 \arcsin(x) \; dx \]
Solution to Example 16
Integration by parts: \( \int w' v \; dx = w v - \int w v' \; dx \).
Let \( w' = x^2 \) hence \( w = (1/3) x^3 \) and \( v = \arcsin(x) \) hence \( v' = \dfrac{1}{\sqrt{1-x^2}} \)
\( \displaystyle \int x^2 \arcsin(x) \; dx = (1/3) x^3 \arcsin(x) - (1/3) \int x^3 \dfrac{1}{\sqrt{1-x^2}} \;dx \qquad (I) \)
The integral on the right may be dealt with substitution: Let \( u = \sqrt {1-x^2} \) which gives \( \dfrac{du}{dx} = -2 x (1/2) (1-x^2)^{-1/2} = - x/u \) or \( \; dx = - (u/x) du \)
\( \displaystyle \int x^3 \dfrac{1}{\sqrt{1-x^2}} \;dx = - \int x^3 \dfrac{1}{u}(u/x) \; du \)
Simplify
\( \displaystyle = - \int x^2 \; du \)
Square both sides of \( u = \sqrt {1-x^2} \) and solve to obtain \( x^2 = 1 - u^2 \) and substitute in the above integral
\( = \displaystyle \int (u^2 - 1) \; du \)
Calculate the above integral
\( = \displaystyle (1/3) u^3 - u + c\)
Substitute back \( u = \sqrt {1-x^2} \) and substitute in integral (I) to obtain the final answer
\[ \displaystyle \int x^2 \arcsin(x) \; dx = \dfrac{1}{3} x^3 \arcsin(x) - \dfrac{1}{3} \left( \dfrac{1}{3} \sqrt {1-x^2}^3 - \sqrt {1-x^2} \right) + c \]
Example 17
Evaluate the integral
\[\displaystyle \int \sqrt x \ln x \; dx \]
Solution to Example 17
Use integration by parts: \( \int w' v \; dx = w v - \int w v' \; dx \). Let \( w' = \sqrt x \) hence \( w = (2/3)x^{3/2} \) and \( v = \ln x \) hence \( v' = 1/x \)
\( \displaystyle \int \sqrt x \ln x \; dx = (2/3)x^{3/2} \ln x - \int (2/3)x^{3/2} (1/x) \; dx \)
Simplify the right side
\( \displaystyle \int \sqrt x \ln x \; dx = (2/3)x^{3/2} \ln x - \int (2/3)x^{1/2} \; dx \)
Evaluate the integral to obtain the final answer
\[ \displaystyle \int \sqrt x \ln x \; dx = \dfrac{2}{3} \; x^{3/2} \ln x - \left(\dfrac{2}{3}\right)^2 x^{3/2} + c \]
Example 18
Evaluate the integral
\[\displaystyle \int \dfrac{\sqrt{x+1}}{x} \; dx \]
Solution to Example 18
Use Integration by Substitution: Let \( u = \sqrt{x+1} \) which gives \( \dfrac{du}{dx} = (1/2) (x+1)^{-(1/2)} \) or \( dx = \dfrac{1}{ (1/2) (x+1)^{-(1/2)} } du = 2 \; u \; du\).
Solve \( u = \sqrt{x+1} \) for \( x \) to obtain \( x = u^2 - 1 \) and rewrite the integral as
\(\displaystyle \int \dfrac{\sqrt{x+1}}{x} \; dx = \int \dfrac{u}{u^2 - 1} 2 u \; du \qquad (I) \)
Simplify
\(\displaystyle \int \dfrac{\sqrt{x+1}}{x} \; dx = 2 \int \dfrac{u^2}{u^2 - 1} \; du \)
Divide the numerator by the denominator in \( \dfrac{u^2}{u^2 - 1} \)
\( \dfrac{u^2}{u^2 - 1} = 1 + \dfrac{1}{u^2-1} \)
and the partial fractions decomposition of \dfrac{1}{u^2-1} helps in rewriting the integrand \( \dfrac{u^2}{u^2 - 1} \) as
\( \dfrac{u^2}{u^2 - 1} = 1 - \dfrac{1}{2\left(u+1\right)}+\dfrac{1}{2\left(u-1\right)} \)
Substitute and calculate the integral in the right side of (I) .
\( \displaystyle 2 \int (1 - \dfrac{1}{2\left(u+1\right)}+\dfrac{1}{2\left(u-1\right)} \; du = 2 (u - \dfrac{1}{2} \ln |u + 1 | + \dfrac{1}{2} \ln |u - 1| ) + c \)
Substitute back \( u = \sqrt{x+1} \) and use properties of the logarithm to obtain the final answer.
\[\displaystyle \int \dfrac{\sqrt{x+1}}{x} \; dx = 2 \sqrt{x+1} + \ln \left(\dfrac{|x|}{(\sqrt{x+1} + 1)^2}\right) + c\]
Example 19
Evaluate the integral
\[\displaystyle \int \sin\left(\sqrt{x}\right) \; dx \]
Solution to Example 19
Use Integration by Substitution: Let \( u = \sqrt{x} \) which gives \( \dfrac{du}{dx} = (1/2) (x)^{-(1/2)} \) or \( dx = \dfrac{1}{ (1/2) (x)^{-(1/2)} } du = 2 \; u \; du\).
\( \displaystyle \int \sin\left(\sqrt{x}\right) \; dx = 2 \int u \sin u \; du \)
Apply integration by parts to \( \int u \sin u \; du \): \( \int w' v \; dx = w v - \int w v' \; dx \). Let \( w' = \sin u \) hence \( w = - \cos u \) and \( v = u \) hence \( v' = 1 \)
\( \displaystyle \int u \sin u \; du = - u \cos u + \int \cos u du = - u \cos u + \sin u + c\)
Back substitute \( u = \sqrt{x} \) to obain the final answer
\[\displaystyle \int \sin\left(\sqrt{x}\right) \; dx = - 2 \sqrt{x} \cos \sqrt{x} +2 \sin \sqrt{x} + c \]
Example 20
Evaluate the integral
\[\displaystyle \int \dfrac{1}{e^x+e^{-x}} \; dx \]
Solution to Example 20
Use Integration by Substitution: Let \( u = e^x \) which gives \( \dfrac{du}{dx} = e^x \) or \( dx = \dfrac{1}{ e^x} du = \dfrac{1}{u} du\).
\(\displaystyle \int \dfrac{1}{e^x+e^{-x}} \; dx = \int \dfrac{1}{u+1/u} \; \dfrac{1}{u} \; du \)
Simplify.
\(\displaystyle = \int \dfrac{1}{u^2+1} \; du \)
Use the common integral \( \int \dfrac{1}{u^2+1}du=\arctan \left(u\right) \) to obtain.
\(\displaystyle = \arctan u + c \)
Substitute back \( u = e^x \) to write the final answer as
\[\displaystyle \int \dfrac{1}{e^x+e^{-x}} \; dx = \arctan e^x + c \]
Example 21
Evaluate the integral
\[\displaystyle \int \log_5 x \; dx \]
Solution to Example 21
We first use the change of base formula of the log: \( \; \log_5 (x) = \dfrac{\ln x }{\ln 5} \) to write the given integral as
\(\displaystyle \int \log_5 x \; dx = \dfrac{1}{\ln 5} \int \ln x \; dx \)
Apply integration by parts to \( \int \ln x \; dx \): \( \int w' v \; dx = w v - \int w v' \; dx \). Let \( w' = 1 \) hence \( w = x \) and \( v = \ln x \) hence \( v' = 1/x\)
\( \displaystyle \int \ln x \; dx = x \ln x - \int x (1/x) \; dx + c\)
Simplify and write the final asnwer as
\[ \displaystyle \int \log_5 x \; dx = \dfrac{1}{\ln 5} ( x \ln x - x) + c \]
Example 22
Evaluate the integral
\[\displaystyle \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx \]
Solution to Example 22
Note that if there is a way to write the expression under the square root as a square, the square root would simplify.
In trigonometry, we have the identity \( \sqrt {1 - \sin^2 t} = \sqrt {\cos^2 t} = | \cos t | \)
We need to make a change of variable such that the expression under the square root in the given integral would be similar to the example above.
\( \sqrt{16 - x^2} = 4 \sqrt {1 - \dfrac{x^2}{16}} = 4 \sqrt {1 - \left(\dfrac{x}{4} \right)^2} \)
Use the trigonometric substitution method: Let \( \; \dfrac{x}{4} = \sin t \) which gives \( x = 4 \sin t \) and \( \dfrac{dx}{dt} = 4 \cos t \) or \( dx = 4 \cos t \; dt \)
\( \displaystyle \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx = \int \dfrac{ (4 \sin t)^2 }{4 | \cos t |} \; 4 \cos t dt \)
The integral above can only be done if we can simplify \( | \cos t | \). For indefinite integrals, we can assume that either \( | \cos t | = \cos t \) or \( | \cos t | = - \cos t \).
For definite integrals we need to study the sign of \( \cos t \) over the interval of integration. The given integral is indefinite and we will assume that \( \cos t \ge 0 \) and therefore \( | \cos t | = \cos t \). Hence the integral may be simplified and written as
\( \displaystyle \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx = 16 \int \sin t^2 \; dt \)
Use the trigonometric identity \( \sin t^2 = (1/2)(1 - \cos (2t) \) to rewrite the integral as
\( \displaystyle \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx = 16 \int (1/2)(1- \cos (2t)t) \; dt \)
Calculate the above integral
\( \displaystyle \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx = 8 t - 4 \sin (2t) +c \)
The substitution \( x = 4 \sin t \) made above may be written as \( t = \arcsin(x/4) \) which is used in the result above and gives the final answer
\[ \displaystyle \int \dfrac{x^2}{\sqrt{16 - x^2}} \; dx = 8 \arcsin(x/4) - 4 \sin (2 \arcsin(x/4)) + c\]
Use the table of integrals and the properties above to calculate the following integrals. [Note that you may need to use more than one of the above properties and methods for one integral].