integration by substitution homework answers

Integration by Substitution

"Integration by Substitution" (also called "u-Substitution" or "The Reverse Chain Rule") is a method to find an integral , but only when it can be set up in a special way.

The first and most vital step is to be able to write our integral in this form:

Like in this example:

When our integral is set up like that, we can do this substitution :

Then we can integrate f(u) , and finish by putting g(x) back as u .

Example: ∫ cos(x 2 ) 2x dx

We know (from above) that it is in the right form to do the substitution:

Now integrate:

∫ cos(u) du = sin(u) + C

And finally put u=x 2 back again:

sin(x 2 ) + C

So ∫ cos(x 2 ) 2x dx = sin(x 2 ) + C

That worked out really nicely! (Well, I knew it would.)

Let's just run through that again in a step-by-step manner:

But this method only works on some integrals of course, and it may need rearranging:

Example: ∫ cos(x 2 ) 6x dx

Oh no! It is 6x , not 2x like before. Our perfect setup is gone.

Never fear! Just rearrange the integral like this:

∫ cos(x 2 ) 6x dx = 3 ∫ cos(x 2 ) 2x dx

(We can pull constant multipliers outside the integration, see Rules of Integration .)

Then go ahead as before:

3 ∫ cos(u) du = 3 sin(u) + C

Now put u=x 2 back again:

3 sin(x 2 ) + C

Now let's try a slightly harder example:

Example: ∫ x/(x 2 +1) dx

Let me see ... the derivative of x 2 +1 is 2x ... so how about we rearrange it like this:

∫ x/(x 2 +1) dx = ½ ∫ 2x/(x 2 +1) dx

Then we have:

Then integrate:

½ ∫ 1/u du = ½ ln | u | + C

Now put u=x 2 +1 back again:

½ ln(x 2 +1) + C

And how about this one:

Example: ∫ (x+1) 3 dx

Let me see ... the derivative of x+1 is ... well it is simply 1.

So we can have this:

∫ (x+1) 3 dx = ∫ (x+1) 3 · 1 dx

∫ u 3 du = u 4 4 + C

Now put u=x+1 back again:

(x+1) 4 4 + C

We can take that idea further like this:

Example: ∫ (5x+2) 7 dx

If it was in THIS form we could do it:

∫ (5x+2) 7 5 dx

So let's make it so by doing this:

1 5 ∫ (5x+2) 7 5 dx

The 1 5 and 5 cancel out so all is fine.

And now we can have u=5x+2

And then integrate:

1 5 ∫ u 7 du = 1 5 u 8 8 + C

Now put u=5x+2 back again, and simplify:

(5x+2) 8 40 + C

• Practice Problems
• Assignment Problems
• Show all Solutions/Steps/ etc.
• Hide all Solutions/Steps/ etc.
• Computing Indefinite Integrals
• More Substitution Rule
• Applications of Derivatives
• Applications of Integrals
• Calculus II
• Calculus III
• Differential Equations
• Algebra & Trig Review
• Common Math Errors
• Complex Number Primer
• How To Study Math
• Cheat Sheets & Tables
• MathJax Help and Configuration
• Notes Downloads
• Complete Book
• Practice Problems Downloads
• Complete Book - Problems Only
• Complete Book - Solutions
• Assignment Problems Downloads
• Other Items
• Get URL's for Download Items
• Print Page in Current Form (Default)
• Show all Solutions/Steps and Print Page
• Hide all Solutions/Steps and Print Page

Section 5.3 : Substitution Rule for Indefinite Integrals

For problems 1 – 16 evaluate the given integral.

• \( \displaystyle \int {{\left( {8x - 12} \right){{\left( {4{x^2} - 12x} \right)}^4}\,dx}}\) Solution
• \( \displaystyle \int {{3{t^{ - 4}}{{\left( {2 + 4{t^{ - 3}}} \right)}^{ - 7}}\,dt}}\) Solution
• \( \displaystyle \int {{\left( {3 - 4w} \right){{\left( {4{w^2} - 6w + 7} \right)}^{10}}\,dw}}\) Solution
• \( \displaystyle \int {{5\left( {z - 4} \right)\,\,\,\sqrt[3]{{{z^2} - 8z}}\,dz}}\) Solution
• \( \displaystyle \int {{90{x^2}\sin \left( {2 + 6{x^3}} \right)\,dx}}\) Solution
• \( \displaystyle \int {{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}}\) Solution
• \( \displaystyle \int {{\left( {15{t^{ - 2}} - 5t} \right)\cos \left( {6{t^{ - 1}} + {t^2}} \right)\,dt}}\) Solution
• \( \displaystyle \int {{\left( {7y - 2{y^3}} \right){{\bf{e}}^{{y^{\,4}} - 7{y^{\,2}}}}\,dy}}\) Solution
• \( \displaystyle \int {{\frac{{4w + 3}}{{4{w^2} + 6w - 1}}\,dw}}\) Solution
• \( \displaystyle \int {{\left( {\cos \left( {3t} \right) - {t^2}} \right){{\left( {\sin \left( {3t} \right) - {t^3}} \right)}^5}\,dt}}\) Solution
• \( \displaystyle \int {{4\left( {\frac{1}{z} - {{\bf{e}}^{ - z}}} \right)}}\cos \left( {{{\bf{e}}^{ - z}} + \ln z} \right)\,dz\) Solution
• \( \displaystyle \int {{{{\sec }^2}\left( v \right){{\bf{e}}^{1 + \tan \left( v \right)}}\,dv}}\) Solution
• \( \displaystyle \int {{10\sin \left( {2x} \right)\cos \left( {2x} \right)\sqrt {{{\cos }^2}\left( {2x} \right) - 5} \,dx}}\) Solution
• \( \displaystyle \int {{\frac{{\csc \left( x \right)\cot \left( x \right)}}{{2 - \csc \left( x \right)}}\,dx}}\) Solution
• \( \displaystyle \int {{\frac{6}{{7 + {y^2}}}\,dy}}\) Solution
• \( \displaystyle \int {{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}}\) Solution
• \( \displaystyle \int {{\frac{{3x}}{{1 + 9{x^2}}}\,dx}}\)
• \( \displaystyle \int {{\frac{{3x}}{{{{\left( {1 + 9{x^2}} \right)}^4}}}\,dx}}\)
• \( \displaystyle \int {{\frac{3}{{1 + 9{x^2}}}\,dx}}\)

Integration by Substitution

Integration by substitution is an important method of integration, which is used when a function to be integrated, is either a complex function or if the direct integration of the function is not feasible. The integral of a function is simplified by this method of integration by substitution , by reducing the given function into a simplified function.

Let us learn the process of integration by substitutions, check some of the important substitutions, and also check the solved examples.

What Is Integration by Substitution?

Integration by substitution is used when the integration of the given function cannot be obtained directly, as the given algebraic function is not in the standard form. Further, the given function can be reduced to the standard form by appropriate substitution. Let us consider the indefinite integral  of a function f(x), \(\int f(x).dx\). Here this integral can be transformed to another form by replacing x with g(t) and by substituting x = g(t).

I = \(\int f(x).dx\)

x = g(t) such that dx/dt = g'(t)

dx = g'(t).dt

I =\(\int f(x).dx = \int f(g(t)).g'(t).dt\)

Steps to Integration by Substitution

The following are the steps that are helpful in performing this method of integration by substitution.

Step - 1: Choose a new variable t for the given function to be reduced.

Step - 2: Determine the value of dx, of the given integral, where f(x) is integrated with respect to x.

Step - 3: Make the required substitution in the function f(x), and the new value dx.

Step - 4: Integrate the function obtained after substitution

Step - 5: Substitute back the initial variable x to obtain the final answer.

Important Substituions in Integration by Substitution

The following are some of the important substitutions which are helpful in simplifying the given expression and easily performing the integration process. Let us check the following specific substitutions for integration by substitution.

• For the integral function \(f(\sqrt{a^2 - x^2})\) we use x = aSinθ or x = aCosθ.
• For the integral of the function \(f(\sqrt {x^2 - a^2})\) we use x = a Secθ or x = aCosecθ.
• For the integral of the function \(f(x^2 + a^2)\). \(f(\sqrt{x^2 + a^2})\) we use x = aTanθ, or x = aCotθ.
• For the integral of the functions \(f(\sqrt {\dfrac{a - x}{a + x}})\) , \(f(\sqrt {\dfrac{a + x}{a - x}})\), we use x = a Cos2θ.

Related Topics

The following topics help in a better understanding of this concept of integration by substitution.

• Integration by Partial Fractions
• Integration by Parts
• Differentiation
• Differential Equations Formula
• Differentiation and Integration Formula

Examples on Integration by Substitution

Example 1: Find the integral of \(\dfrac {e^{Tan^{-1}x}}{1 + x^2}\).

The given expression for integration is \(\int \dfrac {e^{Tan^{-1}x}}{1 + x^2}.dx\)

Here let us take \(Tan^{-1}x = t\), and we can differentiate it further.

\(\dfrac{d}{dx}.Tan^{-1}x = \dfrac{d}{dx}.t\)

\(\dfrac{1}{1 + x^2} = \dfrac{dt}{dx}\)

\(dx =(1 + x^2).dt\)

Let us now substitute the 't' value and dx value in the given expression for integration.

\(\int \dfrac {e^{Tan^{-1}x}}{1 + x^2}.dx = \int \dfrac{e^t}{1+ x^2}.(1 + x^2).dt\)

= \(\int e^t.dt\)

= \(e^t + C\)

= \(e^{Tan^{-1}x} + C\)

Therefore, the integration of \(\dfrac {e^{Tan^{-1}x}}{1 + x^2}\) is \(e^{Tan^{-1}x} + C\).

Example 2: Find the integral of \(2xSec^2(x^2 + 1)\).

The given expression for integration is \(\int 2xSec^2(x^2 + 1).dx\).

Here we substitute \(x^2 + 1 = t\) and differentiate it further.

\(\dfrac{d}{dx}.(x^2 + 1) = \dfrac{d}{dx}.t\)

Let us now substitute t and dx value in the above expression of integration.

\(\int.2xSec^2(x^2 + 1).dx = \int 2xSec^2t.\dfrac{dt}{2x}\)

=\(\int Sec^2t.dt\)

= Tan(x 2 + 1) + C

Therefore the integration of \(\int 2xSec^2(x^2 + 1).dx\) is Tan(x 2 + 1) + C.

go to slide go to slide

Book a Free Trial Class

Practice Questions on Integration by Substitution

Faqs on integration by substitution.

Integration by substitution is an important method of integration, which is used when a function to be integrated, is either a complex function or if the direct integration of the function is not feasible.

How Do You Integrate by Substitution?

Integration by substitution can be performed through a set of sequential steps. First, choose a new variable for the part of the function to be substituted. Secondly, determine the value of differentiation of x, ie dx from this new variable substitution. The third steps involve the process of integration involving this new variable. Finally, substitute back the initial variable to obtain the final answer.

How Do You Know When To Use Integration by Substitution?

The process of integration by substitution is used if the given function to be integrated has one of the following three characteristics.

• The given function has a sub-function.
• The function to be integrated is a complex number -based function.
• The direct integration of the function is not possible.

What Is the Formula For Integration by Substitution?

There is no defined formula for integration by substitution. Based on the given function, the part of the function which is to be substituted is substituted with a new variable.

How Do You Use Integration By Substitution for Trigonometric Formulas?

The integration by substitution is used for trigonometric functions , similar to any other function. Here the trigonometric function is replaced with a new variable, to transform it into an algebraic expression, which is easy to integrate.

10.2 u Substitution Indefinite Integrals

Practice Solutions

Solver Title

Generating PDF...

• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Mean, Median & Mode
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
• Pre Calculus
• First Derivative
• Product Rule
• Quotient Rule
• Sum/Diff Rule
• Second Derivative
• Third Derivative
• Higher Order Derivatives
• Derivative at a point
• Partial Derivative
• Implicit Derivative
• Second Implicit Derivative
• Derivative using Definition
• Slope of Tangent
• Curved Line Slope
• Extreme Points
• Tangent to Conic
• Linear Approximation
• Difference Quotient
• Horizontal Tangent
• One Variable
• Multi Variable Limit
• At Infinity
• L'Hopital's Rule
• Squeeze Theorem
• Substitution
• Sandwich Theorem
• Indefinite Integrals
• Definite Integrals
• Partial Fractions
• U-Substitution
• Trigonometric Substitution
• Weierstrass Substitution
• Long Division
• Improper Integrals
• Antiderivatives
• Double Integrals
• Triple Integrals
• Multiple Integrals
• Limit of Sum
• Area under curve
• Area between curves
• Area under polar curve
• Volume of solid of revolution
• Function Average
• Riemann Sum
• Trapezoidal
• Simpson's Rule
• Midpoint Rule
• Geometric Series Test
• Telescoping Series Test
• Alternating Series Test
• P Series Test
• Divergence Test
• Comparison Test
• Limit Comparison Test
• Integral Test
• Absolute Convergence
• Radius of Convergence
• Interval of Convergence
• Linear First Order
• Linear w/constant coefficients
• Second Order
• Non Homogenous
• System of ODEs
• IVP using Laplace
• Series Solutions
• Method of Frobenius
• Gamma Function
• Taylor Series
• Maclaurin Series
• Fourier Series
• Fourier Transform
• Linear Algebra
• Trigonometry
• Conversions

Most Used Actions

Number line.

• substitution\:\int x^{2}e^{3x}dx
• substitution\:\int\frac{x}{\sqrt{1+x^{2}}}dx
• substitution\:\int 8x\cos(5x)dx,\:u=8x
• substitution\:\int\frac{e^{x}}{e^{x}+e^{-x}}dx,\:u=e^{x}

u-substitution-integration-calculator

• Advanced Math Solutions – Integral Calculator, trigonometric substitution In the previous posts we covered substitution, but standard substitution is not always enough. Integrals involving...

Please add a message.

Message received. Thanks for the feedback.

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

7.1E: Exercises for Integration by Parts

• Last updated
• Save as PDF
• Page ID 18277

In using the technique of integration by parts, you must carefully choose which expression is \(u\). For each of the following problems, use the guidelines in this section to choose \(u\). Do not evaluate the integrals.

1) \(\displaystyle ∫x^3e^{2x}\,dx\)

2) \(\displaystyle ∫x^3\ln(x)\,dx\)

3) \(\displaystyle ∫y^3\cos y\,dy\)

4) \(\displaystyle ∫x^2\arctan x\,dx\)

5) \(\displaystyle ∫e^{3x}\sin(2x)\,dx\)

In exercises 6 - 37, find the integral by using the simplest method. Not all problems require integration by parts.

6) \(\displaystyle ∫v\sin v\,dv\)

7) \(\displaystyle ∫\ln x\,dx\) (Hint: \(\displaystyle ∫\ln x\,dx\) is equivalent to \(\displaystyle ∫1⋅\ln(x)\,dx.)\)

8) \(\displaystyle ∫x\cos x\,dx\)

9) \(\displaystyle ∫\tan^{−1}x\,dx\)

10) \(\displaystyle ∫x^2e^x\,dx\)

11) \(\displaystyle ∫x\sin(2x)\,dx\)

12) \(\displaystyle ∫xe^{4x}\,dx\)

13) \(\displaystyle ∫xe^{−x}\,dx\)

14) \(\displaystyle ∫x\cos 3x\,dx\)

15) \(\displaystyle ∫x^2\cos x\,dx\)

16) \(\displaystyle ∫x\ln x\,dx\)

17) \(\displaystyle ∫\ln(2x+1)\,dx\)

18) \(\displaystyle ∫x^2e^{4x}\,dx\)

19) \(\displaystyle ∫e^x\sin x\,dx\)

20) \(\displaystyle ∫e^x\cos x\,dx\)

21) \(\displaystyle ∫xe^{−x^2}\,dx\)

22) \(\displaystyle ∫x^2e^{−x}\,dx\)

23) \(\displaystyle ∫\sin(\ln(2x))\,dx\)

24) \(\displaystyle ∫\cos(\ln x)\,dx\)

25) \(\displaystyle ∫(\ln x)^2\,dx\)

26) \(\displaystyle ∫\ln(x^2)\,dx\)

27) \(\displaystyle ∫x^2\ln x\,dx\)

28) \(\displaystyle ∫\sin^{−1}x\,dx\)

29) \(\displaystyle ∫\cos^{−1}(2x)\,dx\)

30) \(\displaystyle ∫x\arctan x\,dx\)

31) \(\displaystyle ∫x^2\sin x\,dx\)

32) \(\displaystyle ∫x^3\cos x\,dx\)

33) \(\displaystyle ∫x^3\sin x\,dx\)

34) \(\displaystyle ∫x^3e^x\,dx\)

35) \(\displaystyle ∫x\sec^{−1}x\,dx\)

36) \(\displaystyle ∫x\sec^2x\,dx\)

37) \(\displaystyle ∫x\cosh x\,dx\)

In exercises 38 - 46, compute the definite integrals. Use a graphing utility to confirm your answers.

38) \(\displaystyle ∫^1_{1/e}\ln x\,dx\)

39) \(\displaystyle ∫^1_0xe^{−2x}\,dx\) (Express the answer in exact form.)

40) \(\displaystyle ∫^1_0e^{\sqrt{x}}\,dx \quad (\text{let}\, u=\sqrt{x})\)

41) \(\displaystyle ∫^e_1\ln(x^2)\,dx\)

42) \(\displaystyle ∫^π_0x\cos x\,dx\)

43) \(\displaystyle ∫^π_{−π}x\sin x\,dx\) (Express the answer in exact form.)

44) \(\displaystyle ∫^3_0\ln(x^2+1)\,dx\) (Express the answer in exact form.)

45) \(\displaystyle ∫^{π/2}_0x^2\sin x\,dx\) (Express the answer in exact form.)

46) \(\displaystyle ∫^1_0x5^x\,dx\) (Express the answer using five significant digits.)

47) Evaluate \(\displaystyle ∫\cos x\ln(\sin x)\,dx\)

In exercises 48 - 50, derive the following formulas using the technique of integration by parts. Assume that \(n\) is a positive integer. These formulas are called reduction formulas because the exponent in the \(x\) term has been reduced by one in each case. The second integral is simpler than the original integral.

48) \(\displaystyle ∫x^ne^x\,dx=x^ne^x−n∫x^{n−1}e^x\,dx\)

49) \(\displaystyle ∫x^n\cos x\,dx=x^n\sin x−n∫x^{n−1}\sin x\,dx\)

50) \(\displaystyle ∫x^n\sin x\,dx=\)______

51) Integrate \(\displaystyle ∫2x\sqrt{2x−3}\,dx\) using two methods:

a. Using parts, letting \(dv=\sqrt{2x−3}\,dx\)

b. Substitution, letting \(u=2x−3\)

In exercises 52 - 57, state whether you would use integration by parts to evaluate the integral. If so, identify \(u\) and \(dv\). If not, describe the technique used to perform the integration without actually doing the problem.

52) \(\displaystyle ∫x\ln x\,dx\)

53) \(\displaystyle ∫\frac{\ln^2x}{x}\,dx\)

54) \(\displaystyle ∫xe^x\,dx\)

55) \(\displaystyle ∫xe^{x^2−3}\,dx\)

56) \(\displaystyle ∫x^2\sin x\,dx\)

57) \(\displaystyle ∫x^2\sin(3x^3+2)\,dx\)

In exercises 58-59, sketch the region bounded above by the curve, the \(x\)-axis, and \(x=1\), and find the area of the region. Provide the exact form or round answers to the number of places indicated.

58) \(y=2xe^{−x}\) (Approximate answer to four decimal places.)

59) \(y=e^{−x}\sin(πx)\) (Approximate answer to five decimal places.)

In exercises 60 - 61, find the volume generated by rotating the region bounded by the given curves about the specified line. Express the answers in exact form or approximate to the number of decimal places indicated.

60) \(y=\sin x,\,y=0,\,x=2π,\,x=3π;\) about the \(y\)-axis (Express the answer in exact form.)

61) \(y=e^{−x}, \,y=0,\,x=−1, \, x=0;\) about \(x=1\) (Express the answer in exact form.)

62) A particle moving along a straight line has a velocity of \(v(t)=t^2e^{−t}\) after \(t\) sec. How far does it travel in the first 2 sec? (Assume the units are in feet and express the answer in exact form.)

63) Find the area under the graph of \(y=\sec^3x\) from \(x=0\) to \(x=1\). (Round the answer to two significant digits.)

64) Find the area between \(y=(x−2)e^x\) and the \(x\)-axis from \(x=2\) to \(x=5\). (Express the answer in exact form.)

65) Find the area of the region enclosed by the curve \(y=x\cos x\) and the \(x\)-axis for \(\frac{11π}{2}≤x≤\frac{13π}{2}.\) (Express the answer in exact form.)

66) Find the volume of the solid generated by revolving the region bounded by the curve \(y=\ln x\), the \(x\)-axis, and the vertical line \(x=e^2\) about the \(x\)-axis. (Express the answer in exact form.)

67) Find the volume of the solid generated by revolving the region bounded by the curve \(y=4\cos x\) and the \(x\)-axis, \(\frac{π}{2}≤x≤\frac{3π}{2},\) about the \(x\)-axis. (Express the answer in exact form.)

68) Find the volume of the solid generated by revolving the region in the first quadrant bounded by \(y=e^x\) and the \(x\)-axis, from \(x=0\) to \(x=\ln(7)\), about the \(y\)-axis. (Express the answer in exact form.)

69) What is the volume of the Bundt cake that comes from rotating \( y=\sin x\) around the \(y\)-axis from \( x=0\) to \( x=π\)?