Can you please solve these questions involving integrals? Thanks.

  1. #int 9 cos (5x-3)dx#
  2. #int 4 sin (x/3)dx#
  3. #int (sec^2 7x - csc^2 7x)dx#
  4. #int (sec3xtan3x - 15cos3x)dx#
  5. #int (4csc10xcot10x+2sec^2 10x)dx#

I'm asking for your help in answering these questions for my advanced Math class. I'm okay with integrals in general, but I tend to get confused when it comes to trigonometric functions. Thank you!

2 Answers
Feb 20, 2018

#1.#

#int 9cos(5x-3)dx = 9/5 int cos(5x-3) d(5x-3) = 9/5sin(5x-3) + C#

#2.#

#int 4 sin(x/3)dx = 12 int sin(x/3) d(x/3) = -12cos(x/3) + C#

#3.#

#int (sec^2 7x-csc^2 7x)dx = int (1/(cos^2 7x) - 1/(sin^2 7x))dx#

#int (sec^2 7x-csc^2 7x)dx = int ((sin^2 7x - cos^2 7x) / (cos^2 7x sin^2 7x))dx#

#int (sec^2 7x-csc^2 7x)dx = -4 int (cos 14x) / (sin^2 14x)dx#

#int (sec^2 7x-csc^2 7x)dx = -2/7 int (d(sin 14x)) / (sin^2 14x)#

#int (sec^2 7x-csc^2 7x)dx = 2/(7sin 14 x) +C#

or alternatively:

#int (sec^2 7x-csc^2 7x)dx = 1/7 int sec^2 (7x) d(7x) - 1/7int csc^2(7x)d(7x)#

#int (sec^2 7x-csc^2 7x)dx = (tan (7x) + cot(7x))/7+C#

#4.#

#int (sec 3x tan 3x -15 cos3x)dx = 1/3 int sec 3x tan 3x d(3x) -5 int cos(3x)d(3x)#

#int (sec 3x tan 3x -15 cos3x)dx = sec(3x)/3 -5sin(3x)#

#5.#

#int (4 csc(10x)cot(10x)+2sec^2( 10 x)) dx = 2/5 int csc(10x)cot(10x)d(10x) + 1/5 int sec^2(10x)d(10x)#

#int (4 csc(10x)cot(10x)+2sec^2( 10 x)) dx = (-2csc(10x) +tan(10x))/5#

Feb 20, 2018

Answer:

  1. #9/5sin(5x-3)+C# 2. #-12cos(x/3)#+C 3. #1/7tan(7x)+1/7cot(7x)+C# 4. #1/3sec(3x) - 5sin(3x) + C# 5. #-2/5csc(10x)- 1/5tan(10x)+C#

Explanation:

  1. u=5x-3
    du=5dx
    (1/5)du=dx

Rewrite:
#9/5intcos(u)du# (Bring all constants outside of the integral).
#9/5sin(u)+C#
#9/5sin(5x-3)+C#
2. u=x/3
du=dx/3
3du=dx

Rewrite:
#(4)(3)intsin(u)du#
#12intsin(u)du#
#-12cos(u)+C#
#-12cos(x/3)#+C
3. Split up the integral:
#intsec^2(7x)dx - intcsc^2(7x)dx#

Make the following substitution for both integrals:

u=7x
du=7dx
(1/7)du=dx

#1/7intsec^2(u)du-1/7intcsc^2(u)du#

Now this is only composed of basic integrals:
#1/7tan(u)+1/7cot(u)+C#
#1/7tan(7x)+1/7cot(7x)+C#
4. Split up the integral:
#intsec(3x)tan(3x)dx - int15cos(3x)dx#

Make the following substitution for both integrals:
u=3x
du=3dx
(1/3)du=dx

#1/3intsec(u)tan(u)du - 15/3intcos(u)du#

#1/3intsec(u)tan(u)du - 5intcos(u)du#

Now we have an integral composed of only basic integrals:

#1/3sec(u) - 5sin(u) + C#

#1/3sec(3x) - 5sin(3x) + C#
5. Split up the integral:
#int4csc(10x)cot(10x)dx - int2sec^2(10x)dx#

Make the following substitution for both integrals:

u=10x
du=10dx
(1/10)du=dx
#4/10intcsc(u)cot(u)du - 2/10intsec^2(u)du#

#2/5intcsc(u)cot(u)du - 1/5intsec^2(u)du#

Now this is only composed of basic integrals:

#-2/5csc(u)- 1/5tan(u)+C#

#-2/5csc(10x)- 1/5tan(10x)+C#