Results 1 to 4 of 4

Math Help - trig integrals

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    55

    trig integrals

    i have a calc test coming up and im not sure how to do the following, if you could show me how, it would be greatly appreciated...thanks!

    1. integral (where b= pi/2, a = 0) sin^2x cos^2 x dx

    2. integral (where b= pi/4, a = 0) sec^4 x tan^4 x dx

    3. integral tan^6(ay) dy
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614
    Hello, skabani!


    Here are the first two.
    . . I'll let someone else explain the third one.



    1)\;\;\int^{\frac{\pi}{2}}_0 \sin^2\!x\cos^2\!x\,dx
    We need two identities:
    . . 2\sin\theta\cos\theta \:=\:\sin2\theta\qquad\qquad \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}


    We have: . \sin^2\!x\cos^2\!x \;=\;\frac{1}{4}\left(4\sin^2\!x\cos^2\!x\right) \;=\;\frac{1}{4}(2\sin x\cos x)^2 \;=\;\frac{1}{4}\sin^22x


    The integral becomes: . \frac{1}{4}\int^{\frac{\pi}{2}}_0\sin^22x\,dx \;=\; \frac{1}{4}\int^{\frac{\pi}{2}}_0 \frac{1-\cos4x}{2}\,dx

    And we have: . \frac{1}{8}\int^{\frac{\pi}{2}}_0(1 - \cos4x)\,dx\quad\hdots . Got it?




    2)\;\;\int^{\frac{\pi}{4}}_0 \sec^4\!x\tan^4\!x\,dx

    We have: . \sec^2\!x\cdot\sec^2\!x\cdot\tan^4\!x \;=\;\sec^2\!x\cdot(\tan^2\!x + 1)\cdot\tan^4\!x<br />
\;=\;\sec^2\!x\cdot(\tan^6\!x+\tan^4\!x)

    The integral becomes: . \int^{\frac{\pi}{4}}_0 (\tan^6\!x + \tan^4\!x)\,\sec^2\!x\,dx

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


    Substitute: . \int^b_a(u^6 + u^4)\,du \quad\hdots . Okay?

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2007
    Posts
    55
    great thanks so much!

    i hope someone can help me out with the third one, before tonight, for my exam tomorrow...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    For the last one, first put z=\alpha y, then just worry about \int{\tan ^{6}z\,dz}. Now, note that,

    \tan ^{6}z=\tan ^{4}z\tan ^{2}z=\tan ^{4}z\left( \sec ^{2}z-1 \right)=\tan ^{4}z\sec ^{2}z-\tan ^{4}z.

    Make it in the same fashion for \tan ^{4}z.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig Integrals...
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 18th 2010, 12:10 PM
  2. Trig integrals
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 23rd 2009, 07:38 PM
  3. integrals with trig
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 20th 2009, 12:21 PM
  4. trig integrals
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 24th 2009, 04:43 PM
  5. Trig sub integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 9th 2008, 06:09 PM

Search Tags


/mathhelpforum @mathhelpforum