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Integrals of Trigonometric Functions

Key Questions

  • Recall:

    #int{g'(x)}/{g(x)}dx=ln|g(x)|+C#

    (You can verify this by substitution #u=g(x)#.)

    Now, let us look at the posted antiderivative.

    By the trig identity #tan x={sin x}/{cos x}#,

    #int tan x dx=int{sin x}/{cos x}dx#

    by rewriting it a bit further to fit the form above,

    #=-int{-sin x}/{cos x}dx#

    by the formula above,

    #=-ln|cos x|+C#

    or by #rln x=lnx^r#,

    #=ln|cos x|^{-1}+C=ln|sec x|+C#

    I hope that this was helpful.

  • Since

    #(tanx)'=sec^2x#,

    we have

    #int sec^2x dx=tan x +C#.

    I hope that this was helpful.

  • Since

    #(ln|secx+tanx|)'={secxtanx+sec^2x}/{sec x+tanx}=secx#,

    we have

    #int secx dx=ln|secx+tanx|+C#


    Since

    #(-ln|cscx+cotx|)'=-{-cscxcotx-csc^2x}/{cscx+cotx}=cscx#,

    we have

    #int cscx dx=-ln|cscx+cotx|+C#


    #int cotx dx=int{cosx}/{sinx}dx=ln|sinx|+C#


    I hope that this was helpful.

  • Since

    #(sinx)'=cosx#,

    we have

    #int cosx dx=sinx +C#.


    Since

    #(-cosx)'=sinx#,

    we have

    #int sinx dx=-cosx+C#.


    I hope that this was helpful.

Questions