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Mathematics Dictionary
Dr. K. G. Shih
Integral of trigonometric functions
Subjects

    Symbol Defintion
    Example : Sqr(x) is square root of x
  • Q01 | - ∫sin(x)dx = -cos(x) + C
  • Q02 | - ∫cos(x)dx = +sin(x) + C
  • Q03 | - ∫tan(x)dx = -ln(cos(x)) + C
  • Q04 | - ∫csc(x)dx = ln(csc(x) - cot(x)) + C
  • Q05 | - ∫sec(x)dx = ln(sec(x) + tan(x)) + C
  • Q06 | - ∫cot(x)dx = +ln(sin(x)) + C
  • Q07 | - ∫(sin(x)^2)dx = ?
  • Q08 | - ∫(cos(x)^2)dx = ?
  • Q09 | - ∫(tan(x)^2)dx = ?
  • Q10 | - ∫(csc(x)^2)dx = ?
  • Q11 | - ∫(sec(x)^2)dx = ?
  • Q12 | - ∫(cot(x)^2)dx = ?
  • Q13 | - ∫(sin(x)*cos(x))dx = ?
  • Q14 | - ∫(sin(2*x))dx = ?
Answers


Q01. ∫sin(x)dx = -cos(x) + C

  • Since we know that d/dx(cos(x)) = -sin(x)
  • Hence d(cos(x)) = -sin(x)dx
  • Hence cos(x) = ∫-sin(x)dx
  • Hence ∫sin(x)dx = -cos(x) + C

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Q02. ∫ cos(x)dx = -sin(x) + C

  • Since we know that d/dx(sin(x)) = cos(x)
  • Hence d(sin(x)) = cos(x)dx
  • Hence sin(x) = ∫cos(x)dx
  • Hence ∫cos(x)dx = sin(x) + C

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Q03. ∫ tan(x)dx = -ln(cos(x)) + C

  • Since we know that d/dx(tan(x)) = sec(x)^2
  • Hence d(tan(x)) = (sec(x)^2)dx
  • Hence tan(x) = ∫(sec(x)^2)dx
  • Hence ∫(sec(x)^2)dx = tan(x) + C
  • We can not get intgration of tan(x) by anti-derivative
Porve that integration of tan(x) is -ln(cos(x))
  • tan(x)dx = (sin(x)dx)/cos(x)
  • sin(x)dx = d(-cos(x))
  • Let cos(x) = u, then tan(x)dx = du/u
  • Hence ∫tan(x)dx = -ln(u) + C = -ln(cos(x)) + C
Verify
  • d/dx(-ln(cos(x)) + C) = (-1/cos(x))*(d/dx(cos(x))) = (-1)*(-sin(x)/cos(x) = tan(x)

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Q04. ∫ csc(x)dx = ln(csc(x)) - cot(x))

It is hard to work from left to right. We will find d/dx(F(x)) = csc(x)
  • From right side, we prove d/dx(ln(csc(x) - cot(x)) = csc(x)
    • use chain rule and d/dx(ln(u)) = (1/u)*(du/dx)
    • d/dx(ln(csc(x) - cot(x))
    • = (1/(csc(x) - cot(x))*(d/dx(csc(x) - cot(x))
    • = (1/(csc(x) - cot(x))*(-csc(x)*cot(x) - (-csc(x)^2)
    • = (1/(csc(x) - cot(x))*(csc(x)*(-cot(x) + csc(x))
    • = csc(x)
  • Hence csc(x)dx = d(ln(csc(x) - cot(x))
  • Hence ∫ csc(x)dx = ln(csc(x) - cot(x))

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Q05. ∫ sec(x)dx = ln(sec(x)) + tan(x))

It is hard to work from left to right. We will find d/dx(F(x)) = sec(x)
  • From right side, we prove d/dx(ln(sec(x) + tan(x)) = sec(x)
    • use chain rule and d/dx(ln(u)) = (1/u)*(du/dx)
    • d/dx(ln(sec(x) + tan(x))
    • = (1/(sec(x) + tan(x))*(d/dx(sec(x) + tan(x))
    • = (1/(sec(x) + tan(x))*(sec(x)*tan(x) + sec(x)^2)
    • = (1/(sec(x) + tan(x))*(sec(x)*(tan(x) + sec(x))
    • = sec(x)
  • Hence sec(x)dx = d(ln(sec(x) + tan(x))
  • Hence ∫ sec(x)dx = ln(sec(x) + tan(x))

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Q06. ∫ cot(x)dx = +ln(sin(x)) + C
  • Since we know that d/dx(cot(x)) = -csc(x)^2
  • Hence d(cot(x)) = (-ssc(x)^2)dx
  • Hence cot(x) = ∫(-csc(x)^2)dx
  • Hence ∫(-csc(x)^2)dx = cot(x) + C
  • We can not get intgration of cot(x) by anti-derivative
Porve that integration of cot(x) is ln(sin(x))
  • cot(x)dx = (cos(x)dx)/sin(x)
  • cos(x)dx = d(sin(x))dx
  • Let sin(x) = u, then cot(x)dx = du/u
  • Hence ∫cot(x)dx = ln(u) + C = ln(sin(x)) + C

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Q07. ∫(sin(x)^2)dx = ?

No anti-derivative is given
  • Use cos(2*x) = 1 - 2*sin(x)^2
  • Hence sin(x)^2 = (1 - cos(2*x))/2
  • ∫(sin(x)^2)dx = ∫((1 - cos(2*x)/2)dx = x/2 - sin(2*x)/4
Verify
  • d/dx(x/2 - sin(2*x)/4) = 1/2 - 1/2*cos(2*x) = (1 - cos(2*x))/2 = sin(x)^2

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Q08. ∫(cos(x)^2)dx = ?

No anti-derivative is given
  • Use cos(2*x) = 2*cos(x)^2 - 1
  • Hence cos(x)^2 = (1 + cos(2*x))/2
  • ∫(cos(x)^2)dx = ∫((1 + cos(2*x)/2)dx = x/2 + sin(2*x)/4
Verify
  • d/dx(x/2 + sin(2*x)/4) = 1/2 + 1/2*cos(2*x) = (1 + cos(2*x))/2 = cos(x)^2

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Q09. ∫(tan(x)^2)dx = ?

No anti-derivative is given
  • Use 1 + tan(x)^2 = sec(x)^2 and d/dx(tan(x)) = sec(x)^2
  • Hence tan(x)^2 = sec(2*x)^2 - 1
  • ∫(tan(x)^2)dx = ∫((sec(x)^2 - 1)dx = tan(x) - x
Verify
  • d/dx(tan(x) - x) = sec(x)^2 - 1 = tan(x)^2

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Q10. ∫(csc(x)^2)dx = ?

There is anti-derivative.
  • Since d/dx(cot(x)) = -csc(x)^2
  • Hence d(cot(x)) = (-csc(2*x)^2)dx
  • ∫(-csc(x)^2)dx = cot(x) + C
Verify
  • d/dx(cot(x)) = -csc(x)^2

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Q11. ∫(sec(x)^2)dx = ?

There is anti-derivative.
  • Since d/dx(tan(x)) = sec(x)^2
  • Hence d(tan(x)) = (sec(2*x)^2)dx
  • ∫(sec(x)^2)dx = tan(x) + C
Verify
  • d/dx(tan(x)) = sec(x)^2

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Q12. ∫(cot(x)^2)dx = ?

No anti-derivative is given
  • Use 1 + cot(x)^2 = csc(x)^2 and d/dx(cot(x)) = -csc(x)^2
  • Hence cot(x)^2 = csc(2*x)^2 - 1
  • ∫(cot(x)^2)dx = ∫((csc(x)^2 - 1)dx = -cot(x) - x
Verify
  • d/dx(-cot(x) - x) = csc(x)^2 - 1 = cot(x)^2

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Q13. ∫(sin(x)*cos(x))dx = ?

No anti-derivative is given
  • Let u = sin(x) and then du = cos(x)dx
  • Hence the integrant (sin(x)*cos(x))dx = udu
  • ∫(sin(x)*cos(x))dx = ∫udu = u^2/2 = (sin(x)^2)/2 + C
Verify
  • d/dx((sin(x)^2/)/2) = 2*sin(x)*ccos(x)/2 = sin(x)*cos(x)
Second method : ∫(sin(x)*cos(x))dx = ?

  • Let u = cos(x) and then du = -sin(x)dx
  • Hence the integrant (sin(x)*cos(x))dx = -udu
  • ∫(sin(x)*cos(x))dx = ∫-udu = -u^2/2 = -(cos(x)^2)/2 + C
Verify
  • d/dx((-cos(x)^2/)/2) = -2*cos(x)*(-sin(x))/2 = sin(x)*cos(x)
Formula
  • 1. ∫(sin(x)^n)*cos(x)dx = -(cos(x)^(n+1))/(n+1)
  • 2. ∫(cos(x)^n)*sin(x)dx = +(sin(x)^(n+1))/(n+1)

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Q14. ∫(sin(2*x))dx = ?

No anti-derivative is given
  • Let u = 2*x and du = 2*dx
  • Hence the integant is sin(u)*(1/2)du
  • ∫(sin(2*x))dx = ∫(sin(u)(1/2)du = -cos(u)/2 = -cos(2*x)/2 + C
Verify
  • d/dx(-cos(2*x)/2) = -(-2*sin(2*x))/2 = sin(2*x)
Example : ∫(cos(2*x))dx = ?

No anti-derivative is given
  • Let u = 2*x and du = 2*dx
  • Hence the integant is cos(u)*(1/2)du
  • ∫(cos(2*x))dx = ∫(cos(u)(1/2)du = sin(u)/2 = sin(2*x)/2 + C
Verify
  • d/dx(sin(2*x)/2) = (2*cos(2*x))/2 = cos(2*x)
Formula
  • 1. ∫(sin(n*x))dx = -cos(n*x)/(n+1)
  • 2. ∫(cos(n*x))dx = +sin(n*x)/(n+1)

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