Question

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!

Answer 1

`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`

Answer 2

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`