Results 1 to 3 of 3

Math Help - Trouble with integrating trig functions

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    27

    Trouble with integrating trig functions

    I have two trig functions that i cannot seem to find a way to integrate. They are:

    INTEGRAL of: 31(1+cos(x))^2
    This one looks pretty simple but whenever i check my answer my taking the derivative of it, it never comes out right.

    and

    INTEGRAL of: 27(sin(x)/(cos(x))^3
    Im suppose to have my answer in terms of tangent. Which messes me up because usually i would do a trig substitution to cancel out the sine. If its not possible to express it in terms of tangent then what is it in terms of sine and cosine.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is online now
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by fogel1497 View Post
    I have two trig functions that i cannot seem to find a way to integrate. They are:

    INTEGRAL of: 31(1+cos(x))^2
    This one looks pretty simple but whenever i check my answer my taking the derivative of it, it never comes out right.
    (1+\cos(x))^2=\underbrace{1+2 \cos(x)}_{\text{easy to integrate}}+\cos^2(x)

    Now note that \cos^2(x)=\frac{1+\cos(2x)}{2}

    and

    INTEGRAL of: 27(sin(x)/(cos(x))^3
    Im suppose to have my answer in terms of tangent. Which messes me up because usually i would do a trig substitution to cancel out the sine. If its not possible to express it in terms of tangent then what is it in terms of sine and cosine.
    Substitute u=cos(x)

    get your answer and it'll be ok if you want to transform into tangent, use the identity 1=sinē(x)+cosē(x)
    then transform sin(x)/cos(x)=tan(x)
    but it's not necessary
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    626
    Hello, fogel1497!

    \int \frac{27\sin x}{\cos^3\!x}\,dx

    We have: . 27\int\frac{\sin x}{\cos x}\cdot\frac{1}{\cos^2\!x}\,dx \;=\;27\int \tan x\sec^2\!x\,dx


    Now let: u \:= \:\tan x\quad\Rightarrow\quad du \:= \:\sec^2\!x\,dx


    Got it?


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We could get a different (but equivalent) answer . . .

    We have: . 27\int \sec x \,(\sec x\tan x\,dx)

    Let: u \:=\:\sec x \quad\Rightarrow\quad du \:=\:\sec x\tan x\,dx

    Substitute: . 27\int u\,du \:=\;\frac{27}{2}u^2 + C

    Back-substitute: . \frac{27}{2}\sec^2\!x + C

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integrating trig functions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 5th 2011, 02:16 AM
  2. Integrating trig functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 1st 2009, 01:25 AM
  3. Help Integrating Trig functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 28th 2009, 10:00 AM
  4. [SOLVED] Integrating trig functions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 24th 2008, 06:42 PM
  5. Replies: 3
    Last Post: February 25th 2008, 01:06 PM

Search Tags


/mathhelpforum @mathhelpforum