Calculate the following integrals:

Exercise 1

\int x sin x dx

 

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (26 reviews)
Intasar
£36
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Paolo
4.9
4.9 (13 reviews)
Paolo
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (7 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Farooq
5
5 (14 reviews)
Farooq
£40
/h
1st lesson free!
Tom
5
5 (9 reviews)
Tom
£22
/h
1st lesson free!
Intasar
4.9
4.9 (26 reviews)
Intasar
£36
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Paolo
4.9
4.9 (13 reviews)
Paolo
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (7 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Farooq
5
5 (14 reviews)
Farooq
£40
/h
1st lesson free!
Tom
5
5 (9 reviews)
Tom
£22
/h
First Lesson Free>

Exercise 2

\int \frac{lnx}{x^3}dx

 

Exercise 3

\int (x^3 + 5x^2 - 2) e^{2x} dx

 

Exercise 4

\int ln xdx

 

Exercise 5

\int e^x cos xdx

 

Exercise 6

\int \frac{3x^2 - 2x + 5}{(x + 3)^3} dx

 

Exercise 7

\int x \sqrt{1 + x} dx

 

Exercise 8

\int \frac{dx}{\sqrt{1 + e^x}}

 

Exercise 9

\int \frac{dx}{x \sqrt{x^2 - 2}}

 

Exercise 10

\int csc^3x dx

 

Exercise 11

\int \frac{e ^{4x} + 3} {e ^{3x}}dx

 

 

Solution of exercise 1

\int sin x dx

\int x sin x dx = -x cos x + \int cos x dx = - x cos x + sin x + C

 

Solution of exercise 2

\int \frac{lnx}{x^3} dx

= - \frac{1}{2x^2} lnx - \frac{1}{4x^2} + C

 

Solution of exercise 3

\int (x^3 + 5x^2 - 2) e ^{2x} dx

If u = x^3 + 5x^2 - 2, then u' = 3x^2 + 10x.

If v' = e^{2x} , then v = \frac{1}{2} e^{2x}

= \frac{1}{2} (x^3 + 5x^2 - 2) e^{2x} - \frac{1}{2} \int (3x^2 + 10x) e^ {2x}dx

 

Suppose u = 3x^2 + 10x, then u ' = 6x + 10

If v' = e ^ {2x}, then v = \frac{1}{2} e^{2x}

= \frac{1}{2} (x^3 + 5x^2 - 2) e^{2x} - \frac{1}{2}(\frac{1}{2} (3x^2 + 10 x) e^{2x} - \frac{1}{2} \int (6x + 10)e^{2x} dx)

If u = 6x + 10, then u' = 6

If v' = e^{2x}, then v = \frac{1}{2}e^{2x}

= \frac{1}{2} (x^3 + 5x^2 - 2)e^{2x} - \frac{1}{4} (3x^2 + 10x) e^{2x} + \frac{1}{8} (6x + 10)e^{2x} + \frac{3}{4} \int e^{2x} dx

= \frac{1}{2} (x^3 + 5x^2 - 2)e^{2x} - \frac{1}{4} (3x^2 + 10x) e^{2x} + \frac{1}{8} (6x + 10)e^{2x} - \frac{3}{8}  e^{2x} + C

= (\frac{1}{2} x^3 + \frac{7}{4} x^2 - \frac{7}{4}x - \frac{1}{8}) e^{2x} + C

 

Solution of exercise 4

\int ln x dx

If u = lnx, then u' = \frac{1}{x}

If v' = 1, then v = x

\int lnx dx = x lnx - \int dx = x ln x - x + C

 

Solution of exercise 5

\int e^x cos x dx

If u = e^x, then u' = e^x

If v' = cos x, then v = sin x

\int e^x cos x dx = e^x sin x - \int e^x sin x dx

If u = e^x, then u' = e^x

If v' = sin x, then v = -cos x

= \int e^x cos x dx = e^x sin x + e^x cos x - \int e^x cos x dx

2 \int e^x cos x dx = e^x sin x + e^x cos x

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

 

Solution of exercise 6

\int \frac{3x^2 - 2x + 5}{(x + 3)^3} dx

= \frac{A}{x + 3} + \frac{B}{(x + 3)^2} + \frac{C}{(x + 3)^3}

3x^2 - 2x + 5 = A (x + 3)^2 + B (x + 3) + C

To calculate A, B and C, we substitute x with −3:

Derive and replace with −3:

6x - 2 = 2A (x + 3) + B

x = -3          -20 = B

Derive again:

6 = 2A       A = 3

= \int \frac{3}{x + 3}dx - \int \frac {20}{(x + 3)^2}dx + \int \frac {38}{(x + 3)^3} dx

= 3 ln(x + 3) + \frac{20}{x + 3} - \frac{19}{(x + 3)^2} + C

 

Also, the coefficients can be found by realizing the operations and equaling coefficients:

3x^2 - 2x + 5 = Ax^2 + (6A + B)x + 9A + 3B + C

-2 = 19 + B

5 = 27 + C

3 = A

B = -20

C = 38

 

Solution of exercise 7

\int x \sqrt{1 + x} dx

1 + x = t^2

x = t^2 - 1

dx = 2tdt

\int (t^2 - 1) \cdot t \cdot 2tdt = \int (2t^4 - 2t^2)dt = \frac{2}{5} t^5 - \frac{2}{3}t^3 + C

t = \sqrt{1 + x}

= \frac{2}{5} (\sqrt{1 + x})^5 - \frac{2}{3} (\sqrt{1 + x})^3 + C

= \frac{2}{5} (1 + x)^2 \sqrt{1 + x} - \frac{2}{3} (1 + x) \sqrt{1 + x} + C

 

Solution of exercise 8

\int \frac{dx}{\sqrt{1 + e^x}}

1 + e^x = t^2

e^x = t^2 - 1

e^x dx = 2t dt

dx = \frac{2tdt}{t^2 - 1}

\int \frac{2tdt}{(t^2 - 1)t} = 2 \int \frac{dt}{t^2 - 1}

\frac{1}{t^2 - 1} = \frac{A}{t + 1} + \frac{B}{t - 1}

1 = A(t -1) + B(t + 1)

t = -1       1 = -2A        A = -\frac{1}{2}

t = 1        1 = 2B           B = \frac{1}{2}

2 \int \frac{dt}{t^2 - 1} = - \int \frac{dt}{t + 1} + \int \frac{dt}{t - 1} = - ln(t + 1) + ln (t - 1) + C = ln(\frac{t - 1}{t + 1}) + C

t = \sqrt{1 + e^x}

\int \frac{dx}{\sqrt{1 +e^x}} = ln (\frac{\sqrt{1 + e^x} - 1} {\sqrt{1 + e^x} + 1}) + C

 

Solution of exercise 9

\int \frac{dx}{x \sqrt{x^2 - 2}}

x = \sqrt{2} sect

dx = \sqrt{2}sec t tan t dt

\int \frac{\sqrt{2} sec t tan t}{\sqrt{2} sec t \sqrt{2sec^2 t - 2}} dt = \int \frac{tant}{\sqrt{2 (sec^2t - 1)}}dt = \int \frac{tan t}{\sqrt{2}tant}st

= \frac{1}{\sqrt{2}} \int dt = \frac{1}{\sqrt{2}} + C

x = \sqrt{2} sec t

x = \frac{\sqrt{2}} {cos t}

cos t = \frac{\sqrt{2}} {x}

t = arc cos (\frac{\sqrt{2}} {x})

\int \frac{dx}{x \sqrt{x^2 - 2}} = \frac{1}{\sqrt{2}} arc cos (\frac{\sqrt{2}}{x}) + C

 

Solution of exercise 10

\int csc^3x dx

= \int \frac{dx}{sin^3x}

tan \frac{x}{2} = t

dx = \frac{2dt} {1 + t^2}

\int \frac{1}{(\frac{2t}{1 + t^2})^3} \frac{2dt} {1 + t^2} = \int \frac{(1 +t^2)^2}{4t^3}dx = \int \frac{1 +2t^2 + t^4}{4t^3}dx

= \frac{1}{4} \int \frac{dt}{t^3} + \frac{1}{2} \int \frac {dt}{t} + \frac{1}{4} \int t dt = - \frac{1}{8}t^2 + \frac{1}{2} ln t + \frac{1}{8} t^2 + C

= - \frac{1}{8tan^2 \frac{x}{2}} + \frac{1}{2} ln(tan \frac{x}{2}) + \frac{1}{8} tan^2 \frac{x}{2} + C

 

Solution of exercise 11

\int \frac{e ^{4x} + 3} {e ^{3x}}dx

e^x = t

e^xdx = dt

dx = \frac{dt}{t}

\int \frac{e^{4x} + 3}{e^{3x}} dx = \int \frac{t^4 + 3}{t^3 \cdot t}dt = \int \frac{t^4 + 3}{t^4} dt = \int dt + 3 \int \frac{dt}{t^4} = t - \frac{1}{t^3} + C

= e^x - \frac{1}{e^{3x}} + C

 

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 5.00/5 - 1 vote(s)
Loading...

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.