Results 1 to 7 of 7

Math Help - Integral Multiplication

  1. #1
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267

    Integral Multiplication

    \int tan\;x\sqrt{1+sec^4x}\;dx

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jan 2008
    Posts
    91
    I think you first need to get rid of that square root by doing some substitution.

    Try tan \theta = \sec^2 (x). It might not actually work out...but that 4th power has to be removed before you can replace that \sqrt{1 + (\tan \theta)^2} with just \sec \theta.

    At least that's what I would try. You just shouldn't use integration by parts if it looks like you can still substitute.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by TrevorP View Post
    I think you first need to get rid of that square root by doing some substitution.

    Try tan \theta = \sec^2 (x). It might not actually work out...but that 4th power has to be removed before you can replace that \sqrt{1 + (\tan \theta)^2} with just \sec \theta.

    At least that's what I would try. You just shouldn't use integration by parts if it looks like you can still substitute.
    not sure that would make life any easier, after all, what would \tan x become? \sqrt{\tan \theta - 1}. worst, what would d \theta be? i never wrote anything down, but just thinking about it seems like a mess to me. unless you have some trick to simplify things
    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
    14
    Quote Originally Posted by polymerase View Post
    \int tan\;x\sqrt{1+sec^4x}\;dx
    Okay, this will be a mess. It takes lots of substitutions, but having patience makes sense.

    First rewrite your integral

    \int {\sqrt {1 + \sec ^4 x} \tan x\,dx}  = \int {\sqrt {1 + \cos ^4 x} \sec ^2 x\tan x\,dx} .

    Substitute u=\tan x, the integral becomes

    \int {\frac{{u\sqrt {1 + \left( {u^2  + 1} \right)^2 } }}<br />
{{u^2  + 1}}\,du} .

    Make another substitution according to v=u^2+1, the integral becomes

    \frac{1}<br />
{2}\int {\frac{{\sqrt {1 + v^2 } }}<br />
{v}\,dv} .

    Now a final substitution defined by \varphi ^2  = 1 + v^2. (We're gettin' close.)

    \frac{1}<br />
{2}\int {\frac{{\varphi ^2 }}<br />
{{\varphi ^2  - 1}}\,d\varphi }  = \frac{1}<br />
{2}\left[ {\int {d\varphi }  + \frac{1}<br />
{2}\left( {\int {\frac{1}<br />
{{\varphi  - 1}}\,d\varphi }  - \int {\frac{1}<br />
{{\varphi  + 1}}\,d\varphi } } \right)} \right].

    Now this is easy, so we have

    \frac{1}<br />
{2}\int {\frac{{\varphi ^2 }}<br />
{{\varphi ^2  - 1}}\,d\varphi }  = \frac{1}<br />
{2}\left( {\varphi  + \frac{1}<br />
{2}\ln \left| {\frac{{\varphi  - 1}}<br />
{{\varphi  + 1}}} \right|} \right) + k.

    After back substitute we finally-happily get that

    \int {\sqrt {1 + \sec ^4 x} \tan x\,dx}  = \frac{1}<br />
{2}\left( {\frac{{\sqrt {1 + \cos ^4 x} }}<br />
{{\cos ^2 x}} + \frac{1}<br />
{2}\ln \left| {\frac{{\sqrt {1 + \cos ^4 x}  - \cos ^2 x}}<br />
{{\sqrt {1 + \cos ^4 x}  + \cos ^2 x}}} \right|} \right) \,+ \,k.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2008
    Posts
    91
    \tan \theta = \sec^2 x
    \sec^2 \theta d \theta = 2 \sec x \sec x \tan x dx
    \sec^2 \theta d \theta = 2 \tan \theta \tan x dx
    \frac{\sec^2 \theta d \theta}{2 \tan \theta \tan x} = dx

    \int \frac{\tan x \sqrt{1 + \tan^2 \theta} \sec^2 \theta d \theta }{2 \tan \theta \tan x} = \tfrac{1}{2} \int \frac{\sec^3 \theta d \theta}{\tan \theta }

    Now use integration by parts to solve that.

    EDIT: Ok Krizalid already gave you his answer I guess I'll finish this.

    = \tfrac{1}{2} \int \underbrace{\sec^2 \theta}_{\tfrac{d}{d \theta} \tan \theta} \underbrace{\frac{\sec \theta}{\tan \theta }}_{\csc \theta}  d \theta = \tfrac{1}{2} \Big( \tan \theta \csc \theta - \int \tan \theta (-) \csc \theta \cot \theta d \theta \Big)
    = \tfrac{1}{2} \Big( \sec \theta + \int \csc \theta d \theta \Big) = \tfrac{1}{2} \Big( \sec \theta - ln|\csc \theta + \cot \theta| \Big)
    = \tfrac{1}{2} \Big( \sec ( \arctan ( \sec^2 x ) ) - ln|\csc ( \arctan ( \sec^2 x ) )  + \cot ( \arctan ( \sec^2 x ) )| \Big)
    Last edited by TrevorP; January 30th 2008 at 06:03 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Krizalid View Post
    Okay, this will be a mess. It takes lots of substitutions, but having patience makes sense.

    First rewrite your integral

    \int {\sqrt {1 + \sec ^4 x} \tan x\,dx}  = \int {\sqrt {1 + \cos ^4 x} \sec ^2 x\tan x\,dx} .

    Substitute u=\tan x, the integral becomes

    \int {\frac{{u\sqrt {1 + \left( {u^2  + 1} \right)^2 } }}<br />
{{u^2  + 1}}\,du} .

    Make another substitution according to v=u^2+1, the integral becomes

    \frac{1}<br />
{2}\int {\frac{{\sqrt {1 + v^2 } }}<br />
{v}\,dv} .

    Now a final substitution defined by \varphi ^2  = 1 + v^2. (We're gettin' close.)

    \frac{1}<br />
{2}\int {\frac{{\varphi ^2 }}<br />
{{\varphi ^2  - 1}}\,d\varphi }  = \frac{1}<br />
{2}\left[ {\int {d\varphi }  + \frac{1}<br />
{2}\left( {\int {\frac{1}<br />
{{\varphi  - 1}}\,d\varphi }  - \int {\frac{1}<br />
{{\varphi  + 1}}\,d\varphi } } \right)} \right].

    Now this is easy, so we have

    \frac{1}<br />
{2}\int {\frac{{\varphi ^2 }}<br />
{{\varphi ^2  - 1}}\,d\varphi }  = \frac{1}<br />
{2}\left( {\varphi  + \frac{1}<br />
{2}\ln \left| {\frac{{\varphi  - 1}}<br />
{{\varphi  + 1}}} \right|} \right) + k.

    After back substitute we finally-happily get that

    \int {\sqrt {1 + \sec ^4 x} \tan x\,dx}  = \frac{1}<br />
{2}\left( {\frac{{\sqrt {1 + \cos ^4 x} }}<br />
{{\cos ^2 x}} + \frac{1}<br />
{2}\ln \left| {\frac{{\sqrt {1 + \cos ^4 x}  - \cos ^2 x}}<br />
{{\sqrt {1 + \cos ^4 x}  + \cos ^2 x}}} \right|} \right) + k.
    nice.

    could you have dealt with something like \int \frac {\sqrt{1 + u^4}}u~du in the same way? because we could have gotten that by one substitution, u = \sec x
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by Jhevon View Post
    could you have dealt with something like \int \frac {\sqrt{1 + u^4}}u~du in the same way? because we could have gotten that by one substitution, u = \sec x
    Sure, just substitute v^2=1+u^4\implies v\,dv=2u^3\,du, then

    \int {\frac{{\sqrt {1 + u^4 } }}<br />
{u}\,du}  = \frac{1}<br />
{2}\int {\frac{{v^2 }}<br />
{{v^2  - 1}}\,dv} , and the rest follows.

    (Of course you meant to say u^2=\tan x. )
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Multiplication of 2xy^2. (-3x^2. y)
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 1st 2011, 06:02 PM
  2. [SOLVED] Multiplication
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: February 26th 2011, 10:33 PM
  3. Multiplication
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: December 1st 2010, 10:08 AM
  4. Multiplication mod 2
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 8th 2010, 10:04 AM
  5. Set Multiplication
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: December 6th 2006, 08:16 AM

Search Tags


/mathhelpforum @mathhelpforum