MENU
❌
Home
Math Forum/Help
Problem Solver
Practice
Algebra
Geometry
Tests
College Math
History
Games
MAIN MENU
1 Grade
Adding and subtracting up to 10
Comparing numbers up to 10
Adding and subtracting up to 20
Addition and Subtraction within 20
2 Grade
Adding and Subtracting up to 100
Addition and Subtraction within 20
3 Grade
Addition and Subtraction within 1000
Multiplication up to 5
Multiplication Table
Rounding
Dividing
Addition, Multiplication, Division
Perimeter
4 Grade
Adding and Subtracting
Addition, Multiplication, Division
Equivalent Fractions
Divisibility by 2, 3, 4, 5, 9
Area of Squares and Rectangles
Fractions
Equivalent Fractions
Least Common Multiple
Adding and Subtracting
Fraction Multiplication and Division
Operations
Mixed Numbers
Decimals
Expressions
6 Grade
Percents
Signed Numbers
The Coordinate Plane
Equations
Expressions
Polynomials
Symplifying Expressions
Polynomial Vocabulary
Polynomial Expressions
Factoring
7 Grade
Angles
Inequalities
Linear Functions
8 Grade
Congurence of Triangles
Linear Functions
Systems of equations
Slope
Parametric Linear Equations
Word Problems
Exponents
Roots
Quadratic Equations
Quadratic Inequalities
Rational Inequalities
Vieta's Formulas
Progressions
Arithmetic Progressions
Geometric Progression
Progressions
Number Sequences
Reciprocal Equations
Logarithms
Logarithmic Expressions
Logarithmic Equations
Trigonometry
Trigonometry
Identities
Trigonometry
Trigonometric Inequalities
Extremal value problems
Numbers Classification
Geometry
Intercept Theorem
Slope
Law of Sines
Law of Cosines
Vectors
Probability
Limits of Functions
Properties of Triangles
Pythagorean Theorem
Matrices
Complex Numbers
Inverse Trigonometric Functions
Analytic Geometry
Analytic Geometry
Circle
Parabola
Ellipse
Conic sections
Polar coordinates
Integrals
Integrals
Integration by Parts
Trigonometric Substitutions
Application
Differential Equations
Home
Practice
Integration by Parts
Integration by Parts: Problems with Solutions
By Prof. Hernando Guzman Jaimes (University of Zulia - Maracaibo, Venezuela)
Problem 1
Evalutate the integral $\int x^{3}\ln\ x\ dx$, using integration by parts.
Let
$u=\ln x,$
$dv=x^{3}dx.$
$\int x^{3}\ln $ $x$ $dx=x^{3}\left( \frac{1}{4}\ln x+\frac{1}{4} \right) +C$
$\int x^{3}\ln $ $x$ $dx=x^{2}\left( \frac{1}{4}\ln x-\frac{1}{4} \right) +C$
$\int x^{3}\ln $ $x$ $dx=x^{3}\left( \frac{1}{4}\ln x-\frac{1}{4} \right) +C$
$\int x^{3}\ln $ $x$ $dx=x^{4}\left( \frac{1}{4}\ln x-\frac{1}{4} \right) +C$
Solution:
We have to find $\int x^{3}\ln $ $x$ $dx$
$u=\ln $ $x,$ $dv=x^{3}dx.$
Then $du=\frac{dx}{x}\qquad v=\int x^{3}dx=\frac{x^{4}}{4}$ and now we apply
$\int udv=uv-\int vdu=\frac{x^{4}}{4}\ln $ $x-\int \frac{x^{4}}{4}\frac{dx}{x}=\frac{x^{4}}{4}\ln $ $x-\frac{1}{4}\int x^{3}dx$
So $\int x^{3}\ln $ $x$ $dx=\frac{x^{4}}{4}\ln $ $x-\frac{1}{16}x^{4}+C=$
$x^{4}\left( \frac{1}{4}\ln x-\frac{1}{4}\right) +C$
Problem 2
Let $u=4x+7,$ $dv=e^{x}dx$
Evaluate the integral $\int (4x+7)e^{x}dx$ using integration by parts.
$\int (4x+7)e^{x}dx=\left( 4x-3\right) e^{x}+C$
$\int (4x+7)e^{x}dx=\left( 4x+3\right) e^{2x}+C$
$\int (4x+7)e^{x}dx=\left( 4x+3\right) e^{x}+C$
$\int (4x+7)e^{x}dx=\left( 4x^{2}+3\right) e^{x}+C$
Solution:
If $u=4x+7,$ $dv=e^{x}dx$ then $du=4\qquad v=\int e^{x}dx=e^{x}$
so $\int udv=uv-\int vdu$ then $\int (4x+7)e^{x}dx=\left( 4x+7\right) e^{x}-4\int e^{x}dx$
$\int (4x+7)e^{x}dx=\left( 4x+7\right) e^{x}-4e^{x}+C=\left( 4x+3\right) e^{x}+C$
Problem 3
Evalutate the integral $\int x\sin 3x$ $dx$ using intregation by parts.
Consider $u=x,$
$dv=\sin 3x$ $dx$.
$\int x\sin 3x\ dx =-\frac{1}{3}x\cos 3x+\frac{1}{9} \sin 3x+C$
$\int x\sin 3x\ dx =-\frac{1}{3}x\sin 3x+\frac{1}{9} \cos 3x+C$
$\int x\sin 3x\ dx =-\frac{1}{3}x\sin 3x+\frac{1}{9} \sin 3x+C$
$\int x\sin 3x\ dx =-\frac{1}{3}x\cos 3x+\frac{1}{9} \cos 3x+C$
Solution:
Let $u=x\qquad dv=\sin 3x$ $dx$ then
$du=dx\qquad v=\int \sin 3x$ $dx$
$v=-\frac{1}{3}\cos 3x$ so $\int udv=uv-\int vdu$ and replacing
$\int x\sin 3x$ $dx=-\frac{1}{3}x\cos 3x-\int -\frac{1}{3}\cos 3xdx=$
$=-\frac{1}{3}x\cos 3x+\frac{1}{3}\int \cos 3xdx=-\frac{1}{3}x\cos 3x+\frac{1}{9}\sin 3x+C$
Problem 4
Using integration by parts find the integral $\int x\cos 4x\ dx$,
Consider $ u=x,$
$dv=\cos 4x$ $dx.$
$\int x\cos 4x\ dx=\frac{1}{4}x\sin 4x+\frac{1}{16}\cos x+C$
$\int x\cos 4x\ dx=\frac{1}{4}x\sin x+\frac{1}{16}\cos 4x+C$
$\int x\cos 4x\ dx=\frac{1}{4}x\sin 4x+\frac{1}{16}\cos 4x+C$
$\int x\cos 4x\ dx=\frac{1}{4}x\sin x+\frac{1}{16}\cos x+C$
Solution:
When we do $u=x,$ $dv=\cos 4x\ dx$ we obtain that
$du=dx\qquad v=\int \cos 4x$ $dx=\frac{1}{4}\sin 4x$
then if we use the formula $\int udv=uv-\int vdu$
$\int x\cos 4x\ dx=\frac{1}{4}x\sin 4x-\frac{1}{4}\int \sin 4xdx$ then
$\int x\cos 4x\ dx=\frac{1}{4}x\sin 4x+\frac{1}{16}\cos 4x+C$
Problem 5
Using integration by parts find the integral
$\int \cos ^{3}x\sin x\ dx$.
$\int \cos ^{3}x\sin x\ dx=\frac{\sin ^{4}x}{2}-\frac{\sin ^{2}x}{4}+C$
$\int \cos ^{3}x\sin x\ dx=\frac{\sin ^{2}x}{2}-\frac{\sin ^{4}x}{4}+C$
$\int \cos ^{3}x\sin x\ dx=\frac{\cos ^{2}x}{2}-\frac{\sin ^{4}x}{4}+C$
$\int \cos ^{3}x\sin x\ dx=\frac{\sin ^{2}x}{2}-\frac{\cos ^{4}x}{4}+C$
Solution:
$\int \cos ^{3}x\sin x$ $dx=\int \cos ^{2}x\sin x\cos x$ $dx$ and now we do
$\cos ^{2}x=1-\sin ^{2}x$ so $\int \cos ^{3}x\sin x$ $dx=$
$=\int \left( 1-\sin ^{2}x\right) \sin x\cos x$ $dx=\int \left( \sin x-\sin ^{3}x\right) \cos x$ $dx$
and now $u=\sin x\Rightarrow du=\cos xdx$ then
$\int \cos ^{3}x\sin x$ $dx=\int udu-\int u^{3}du=\frac{u^{2}}{2}-\frac{u^{4}}{4}+C$
then $\int \cos ^{3}x\sin x$ $dx=\frac{\sin ^{2}x}{2}-\frac{\sin ^{4}x}{4}+C$
Problem 6
Using integration by parts find the integral:
$\int \cos ^{3}x\sin ^{4}x\ dx$
$\int \cos ^{3}x\sin ^{4}x\ dx=\frac{\sin ^{5}x}{5}-\frac{\cos ^{7}x}{7}+C$
$\int \cos ^{3}x\sin ^{4}x\ dx=\frac{\cos ^{5}x}{5}-\frac{\sin ^{7}x}{7}+C$
$\int \cos ^{3}x\sin ^{4}x\ dx=\frac{\sin ^{7}x}{5}-\frac{\sin ^{5}x}{7}+C$
$\int \cos ^{3}x\sin ^{4}x\ dx=\frac{\sin ^{5}x}{5}-\frac{\sin ^{7}x}{7}+C$
Solution:
$\int \cos^{3}x\sin^{4}x\ dx=\int \cos^{2}x\sin^{4}x\cos x$ $dx$ and now we do
$\cos ^{2}x=1-\sin ^{2}x$
So $\int \cos^{3}x\sin^{4}x$ $dx=$
$=\int \left( 1-\sin ^{2}x\right) \sin ^{4}x\cos x$ $dx=\int \left( \sin^{4}x-\sin ^{6}x\right) \cos x\ dx$
and now $u=\sin x\Rightarrow du=\cos xdx$ then
$\int \cos ^{3}x\sin ^{4}x$ $dx=\int u^{4}du-\int u^{6}du=\frac{u^{5}}{5}-\frac{u^{7}}{7}+C$
then $\int \cos ^{3}x\sin ^{4}x$ $dx=\frac{\sin ^{5}x}{5}-\frac{\sin ^{7}x}{7}+C$
Problem 7
Evalutate the following integral
$\int \sin ^{3}x\cos ^{2}x\ dx$
$\int \sin ^{3}x\cos ^{2}x\ dx=-\frac{\cos ^{3}x}{3}+\frac{\cos ^{5}x}{5}+C$
$\int \sin ^{3}x\cos ^{2}x\ dx=-\frac{\sin ^{3}x}{3}+\frac{\cos ^{5}x}{5}+C$
$\int \sin ^{3}x\cos ^{2}x\ dx=-\frac{\cos ^{3}x}{3}+\frac{\sin ^{5}x}{5}+C$
$\int \sin ^{3}x\cos ^{2}x\ dx=-\frac{\sin ^{3}x}{3}+\frac{\sin ^{5}x}{5}+C$
Solution:
$\int \sin ^{3}x\cos ^{2}x\ dx=\int \sin ^{2}x\sin x\cos ^{2}x\ dx$ and since
$\sin ^{2}x=1-\cos ^{2}x$ we get $\int \sin ^{3}x\cos ^{2}x\ dx=$
$=\int \left( 1-\cos ^{2}x\right) \sin x\cos ^{2}xdx=\int \left( \cos^{2}x-\cos ^{4}x\right) \sin xdx$
Now we can do $u=\cos x\qquad du=-\sin xdx$ so
$\int \sin ^{3}x\cos ^{2}x$ $dx=-\int u^{2}du+\int u^{4}du=-\frac{u^{3}}{3}+ \frac{u^{5}}{5}+C$
then $\int \sin ^{3}x\cos ^{2}x$ $dx=-\frac{\cos ^{3}x}{3}+\frac{\cos ^{5}x}{5}+C$
Problem 8
$\int \sin^{3}x\ dx=$
$\int \sin ^{3}x\ dx=-\cos x+\frac{\cos ^{2}x}{2}+C$
$\int \sin ^{3}x\ dx=-\cos x+\frac{\sin ^{3}x}{3}+C$
$\int \sin ^{3}x\ dx=-\cos x+\frac{\cos ^{3}x}{3}+C$
$\int \sin ^{3}x\ dx=-\sin x+\frac{\cos ^{3}x}{3}+C$
Solution:
We can do, $\int \sin ^{3}x\ dx=\int \sin ^{2}x\sin x$ $dx=\int \left( 1-\cos ^{2}x\right) \sin x\ dx$
then $\int \sin ^{3}x$ $dx=\int \sin xdx-\int \cos ^{2}x\sin x\ dx$ and
substitute $u=\cos x\qquad du=-\sin xdx$ so
$\int \sin ^{3}x$ $dx=-\cos x+\int u^{2}du=-\cos x+\frac{u^{3}}{3}+C$
Then $\int \sin ^{3}x$ $dx=$ $-\cos x+\frac{\cos ^{3}x}{3}+C$
Problem 9
Evaluate the following integral
$\int xe^{-2x}dx$
using the method of integration by parts.
$\int xe^{-2x}\ dx=-\frac{1}{2}xe^{-2x}+\frac{1}{4}e^{-3x}+C$
$\int xe^{-2x}\ dx=\frac{1}{2}xe^{-2x}+\frac{1}{4}e^{-2x}+C$
$\int xe^{-2x}\ dx=-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C$
$\int xe^{-2x}\ dx=-\frac{1}{2}xe^{-x}-\frac{1}{4}e^{-x}+C$
Solution:
Substitute $u=x\qquad dv=e^{-2x}\ dx$
So $du=dx\qquad v=\int e^{-2x}$ $dx=-\frac{1}{2}e^{-2x}$ then we apply
$\int udv=uv-\int vdu\Rightarrow$
$\int xe^{-2x}$ $dx=-\frac{1}{2}xe^{-2x}-\int -\frac{1}{2}e^{-2x}dx$
So $\int xe^{-2x}$ $dx=-\frac{1}{2}xe^{-2x}+\frac{1}{2}\int e^{-2x}dx= -\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C$
Problem 10
$\int \frac{2x}{e^{x}}\ dx =$
$\int \frac{2x}{e^{x}}$ $dx=e^{-x}\left( 2x-2\right) +C$
$\int \frac{2x}{e^{x}}$ $dx=-e^{-x}\left( 2x^{2}+2\right) +C$
$\int \frac{2x}{e^{x}}$ $dx=e^{x}\left( 2x+2\right) +C$
$\int \frac{2x}{e^{x}}$ $dx=-e^{-x}\left( 2x+2\right) +C$
Solution:
Rewrite the integral as follows $\int \frac{2x}{e^{x}}\ dx=2\int xe^{-x}dx$
Then apply integration by parts method
$\int udv=uv-\int vdu$ with $u=x$
$dv=e^{-x}dx\Rightarrow du=dx$
$v=\int e^{-x}dx=-e^{-x}$ and now, we substitute in
$\int \frac{2x}{e^{x}}$ $dx=2\int xe^{-x}dx=2\left( -xe^{-x}-\int -e^{-x}dx\right) =-2xe^{-x}-2e^{-x}+C$
$\int \frac{2x}{e^{x}}$ $dx=-e^{-x}\left( 2x+2\right) +C$
Problem 11
$\int x^{2}e^{x}\ dx = $
$\int x^{2}e^{x}$ $dx=e^{x}\left( x^{2}+2x+2\right) +C$
$\int x^{2}e^{x}$ $dx=e^{x}\left( x^{2}-2x+2\right) +C$
$\int x^{2}e^{x}$ $dx=e^{x}x^{2}+C$
$\int x^{2}e^{x}$ $dx=e^{x}x^{3}+C$
Solution:
In this case, we apply twice the method of integration by parts, to reduce the degree of the power of $x$, from $2,1,0$
and simplify the integral
$u=x^{2}\qquad dv=e^{x}$ $dx$
then $du=2xdx\qquad v=\int e^{x}$ $dx=e^{x}$ and we replace in the formula
$\int udv=uv-\int vdu$ so,
$\int x^{2}e^{x}$ $dx=x^{2}e^{x}-2$ $\int xe^{x}dx$
We use integration by parts:
$\int xe^{x}dx$ with
$u=x\qquad dv=e^{x}$ $dx$
$u=dx\qquad v=\int e^{x}$ $dx=e^{x}$
then $\int x^{2}e^{x}$ $dx=x^{2}e^{x}-2\left( xe^{x}-\int e^{x}dx\right) =x^{2}e^{x}-2xe^{x}+2e^{x}+C$
So $\int x^{2}e^{x}$ $dx=e^{x}\left( x^{2}-2x+2\right) +C$
Problem 12
$\int t$ $\ln (t+1)$ $dt=$
$\ln (t+1)\left( \frac{1}{2}t^{2}-\frac{1}{2}\right) +C$
$-\frac{1}{2}\left( \frac{\left(t+1\right) ^{2}}{2}-2\left( t+1\right) \right) +C$
$\ln (t+1)\left( \frac{1}{2}t^{2}\right)-\frac{1}{2}+C$
$\ln(t+1)\left(\frac12t^2-\frac12\right)-\frac12\left(\frac{(t+1)^2}{2}-2(t+1)\right)+C$
Solution:
Here the substitution is $u=\ln (t+1)\qquad dv=tdt$ so
$du=\frac{1}{t+1}dt\qquad v=\int tdt=\frac{1}{2}t^{2}$ and replacing in the formula
$\int udv=uv-\int vdu$ we obtain
$\int t\ \ln (t+1)\ dt=\frac{1}{2}t^{2}\ln (t+1)-\frac{1}{2}$
$\int \frac{t^{2}}{t+1}dt$ Now to solve this
Substitute $z=t+1$
then $t=z-1$ and $dz=dt$
So
$\int \frac{t^{2}}{t+1}dt=\int \frac{\left(z-1\right)^{2}}{z}dz$
$=\int \frac{z^{2}-2z+1}{z}dz=\int zdz-2\int dz+\int z^{-1}dz=\frac{z^{2}}{2}-2z+\ln (z)$
Since $z=t+1$
$\int \frac{t^{2}}{t+1}dt=\frac{\left( t+1\right)^{2}}{2}-2\left( t+1\right) +\ln (t+1)$ so $\int t\ln (t+1)dt=\frac{1}{2}t^{2}\ln (t+1)-\frac{1}{2}\left[ \frac{\left( t+1\right) ^{2}}{2}-2\left( t+1\right) +\ln (t+1)\right] +C$
$\int t$ $\ln (t+1)\ dt=\ln (t+1)\left[ \frac{1}{2}t^{2}-\frac{1}{2}\right] -\frac{1}{2}\left[ \frac{\left( t+1\right) ^{2}}{2}-2\left( t+1\right)\right] +C$
Problem 13
$\int \frac{\ln (x)}{x^{2}}dx= $
$\int \frac{\ln (x)}{x^{2}}dx=\frac{1}{x^{2}}\left( \ln x+1\right) +C$
$\int \frac{\ln (x)}{x^{2}}dx=-\frac{1}{x}\left( \ln x+1\right) +C$
$\int \frac{\ln (x)}{x^{2}}dx=\frac{1}{x}\left( \ln x-1\right) +C$
None of these.
Solution:
In this case we do, $u=\ln (x)\qquad dv=x^{-2}dx$ then
$du=\frac{dx}{x}\qquad v=\int x^{-2}dx=-x^{-1}$ and now we replace in
$\int udv=uv-\int vdu$ then $\int \frac{\ln (x)}{x^{2}}dx=-\frac{\ln (x)}{x}+\int x^{-2}dx$ so
$\int \frac{\ln (x)}{x^{2}}dx=-\frac{\ln (x)}{x}-x^{-1}+C=-\frac{1}{x}\left( \ln x+1\right) +C$
Problem 14
Evaluate the integral using integration by parts.
$\int x\cos x$ $dx$
$\int x\cos x\ dx=x\sin x+\cos x+C$
$\int x\cos x\ dx=x\sin x-\cos x+C$
$\int x\cos x\ dx=x\sin x+x\cos x+C$
$\int x\cos x\ dx=x\cos x+\sin x+C$
Solution:
We have to do $u=x\qquad dv=\cos x$ $dx$ so $du=dx$ and
$v=\int \cos x\ dx=\sin x$ then if we apply the formula
$\int udv=uv-\int vdu$ we obtain
$\int x\cos x\ dx=x\sin x-\int \sin xdx$
So $\int x\cos x$ $dx=x\sin x+\cos x+C$
Problem 15
Evaluate the integral $\int x\sin x\ dx$ using the method of integration by parts.
$\int x\sin x\ dx=-x\cos x+\sin x+C$
$\int x\sin x\ dx=-x\sin x+\cos x+C$
$\int x\sin x\ dx=x\cos x-\sin x+C$
$\int x\sin x\ dx=-x^{2}+\cos x+\sin x+C$
Solution:
Let $u=x\qquad dv=\sin xdx$ then $du=dx$ and $v=\int \sin xdx$
$v=-\cos x$, now we substitute in $\int udv=uv-\int vdu$ so
$\int x\sin x$ $dx=-x\cos x+\int \cos xdx=-x\cos x+\sin x+C$
Problem 16
$\int x^{2}\sin x$ $dx=$
$x^{2}\cos x+2\left( x\sin x+\cos x\right) +C$
$-x\sin x-2\left( x\sin x+\cos x\right) +C$
$-x^{2}\cos x-2\left( x\sin x+\cos x\right) +C$
$-x^{2}\cos x-2\left( x\cos x+\sin x\right) +C$
Solution:
In this case, we must apply twice the method of integration
by parts, to reduce the degree of the power of $x$, from $2,1,0$
and simplify the integral, so we do,
$u=x^{2}\qquad dv=\sin x\ dx$
then $du=2xdx\qquad v=-\cos x$ then
$\int udv=uv-\int vdu$
$\int x^{2}\sin x$ $dx=-x^{2}\cos x-2\int x\cos xdx$
Now we apply integration by parts to the integral
$\int x\cos xdx=x\sin x-\int \sin xdx$
Let $u=x\qquad dv=\cos xdx$ and $du=dx\qquad v=\sin x$
So, $\int x\cos xdx=x\sin x+\cos x$ then $\int x^{2}\sin x\ dx=$
$=-x^{2}\cos x-2\left( x\sin x+\cos x\right) +C$
Problem 17
Determine the solution of the integral
$\int e^{2x}\sin x\ dx=$
$-\frac{1}{3}e^{2x}\sin x+C$
$-\frac{1}{3}e^{2x}\cos x+C$
$\frac{1}{3}e^{3x}\cos x+C$
$-\frac{1}{3}e^{3x}\tan x+C$
Solution:
Here we can find a recursive solution if $I=\int e^{2x}\sin x\ dx$
and we apply integration by parts, then $u=e^{2x},dv=\sin x\ dx$
$du=2e^{2x}dx\qquad v=-\cos x$
so $\int udv=uv-\int vdu\Longrightarrow \int e^{2x}\sin x\ dx=-e^{2x}\cos x+2\int \cos xe^{2x}dx$
now again we apply integration by parts,
so $u=e^{2x},dv=\cos x\ dx$
Then $du=2e^{2x}dx\qquad v=\sin x$
And replacing in the integral
$I=\int e^{2x}\sin x\ dx=-e^{2x}\cos x+2 \int \cos xe^{2x}dx$ now we apply integration by parts
$u=e^{2x}\qquad dv=\sin xdx$
$du=2e^{2x}dx\qquad v=-\cos x$
so, $I=\int e^{2x}\sin x\ dx=-e^{2x}\cos x+2\int -e^{2x}\cos xdx$
$=-e^{2x}\cos x-2\int e^{2x}\cos xdx=-e^{2x}\cos x-2I$ then
$3I=-e^{2x}\cos x+C\Rightarrow I=-\frac{1}{3}e^{2x}\cos x+C$
We get $I=\int e^{2x}\sin x\ dx=-\frac{1}{3}e^{2x}\cos x+C$
Problem 18
$I=\int e^{-3x}\sin 5x\ dx = $
$I=\frac{225}{226}\left( -\frac{1}{5}e^{3x}\sin 5x-\frac{1}{75}e^{-3x}\sin 5x\right) +C$
$I=\frac{225}{226}\left( -\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{75}e^{-3x}\cos 5x\right) +C$
$I=\frac{225}{226}\left( -\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{75}e^{-3x}\sin 5x\right) +C$
$I=\frac{225}{226}-\frac{1}{5}e^{-3x}\tan 5x+C$
Solution:
$\int e^{-3x}\sin 5x$ $dx\qquad $ in this case we do
$u=e^{-3x}\qquad dv=\sin 5x\ dx$
$du=-\frac{1}{3}e^{-3x}dx\qquad v=-\frac{1}{5}\cos 5x$
Since $I=\int e^{-3x}\sin 5x\ dx$
$I=uv-vdu=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{15}\int e^{-3x}\cos 5xdx$
We apply integration by parts to this integral again, so
$u=e^{-3x}\qquad dv=\cos 5x$ $dx\Longrightarrow du=-\frac{1}{3}e^{-3x}dx\qquad v=\frac{1}{5}\sin 5x$
Then $I=uv-vdu=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{15}\int e^{-3x}\cos 5xdx$
So $I=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{15}\left[ \frac{1}{5}e^{-3x}\sin 5x+\frac{1}{15}\int e^{-3x}\sin 5xdx\right] $
then $I+\frac{1}{225}I=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{75}e^{-3x}\sin 5x$
and now
$\frac{226}{225}I=-\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{75}e^{-3x}\sin 5x$ so
$I=\frac{225}{226}\left( -\frac{1}{5}e^{-3x}\cos 5x-\frac{1}{75}e^{-3x}\sin 5x\right) +C$
Problem 19
Determine the solution of the following integral
$\int \sec ^{2}x\tan x\ dx$
$\int \sec ^{2}x\tan x\ dx=\frac{\tan ^{2}x}{2}+C$
$\int \sec ^{2}x\tan x\ dx=\frac{\sec ^{2}x}{2}+C$
$\int \sec ^{2}x\tan x\ dx=\frac{\csc ^{2}x}{2}+C$
$\int \sec ^{2}x\tan x\ dx=\frac{\sin ^{2}x}{2}+C$
Solution:
If $u=\tan x\Rightarrow du=\sec ^{2}xdx$ and
$\int \sec ^{2}x\tan x\ dx=\int udu$
So $\int \sec ^{2}x\tan x$ $dx=\frac{u^{2}}{2}+C=\frac{\tan ^{2}x}{2}+C$
Problem 20
Solve the following trigonometric integral
$\int \tan ^{2}x\sec ^{4}x\ dx$
$\int \tan ^{2}x\sec ^{4}x\ dx=\frac{\tan 2x}{5}+\frac{\tan^{4}x}{3}+C$
$\int \tan ^{2}x\sec ^{4}x\ dx=\frac{\sec ^{5}x}{5}+\frac{\sec^{3}x}{3}+C$
$\int \tan ^{2}x\sec ^{4}x\ dx=\frac{\cos ^{5}x}{5}+\frac{\cos^{3}x}{3}+C$
$\int \tan ^{2}x\sec ^{4}x\ dx=\frac{\tan ^{5}x}{5}+\frac{\tan^{3}x}{3}+C$
Solution:
We do $\int \tan ^{2}x\sec^{2}x\sec^{2}x\ dx$ and since
$\sec^{2}x=\tan ^{2}x+1$
We get $\int \tan^{2}x\sec^{4}x\ dx=\ \int \tan ^{2}x\left( \tan^{2}x+1\right) \sec ^{2}x\ dx$
$=\int \left( \tan ^{4}x+\tan^{2}x\right) \sec ^{2}x$ $dx$
We can do now:
If $u=\tan x\Rightarrow du=\sec^{2}xdx$ so $\int \tan^{2}x\sec^{4}x\ dx=$
$=\int \left( u^{4}+u^{2}\right) du=\frac{u^{5}}{5}+\frac{u^{3}}{3}+C$ then
$\int \tan^{2}x\sec^{4}x\ dx=\frac{\tan ^{5}x}{5}+\frac{\tan ^{3}x}{3}+C$
Submit a problem on this page.
Problem text:
Solution:
Answer:
Your name(if you would like to be published):
E-mail(you will be notified when the problem is published)
Notes
: use [tex][/tex] (as in the forum if you would like to use latex).
Correct:
Wrong:
Unsolved problems:
Contact email:
Follow us on
Twitter
Facebook
Copyright © 2005 - 2021.