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Math Help - More integrals

  1. #1
    Senior Member polymerase's Avatar
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    More integrals

    \int \frac{\sec\:x\;\cos\:2x}{\sin\;x+\sec\;x}\;dx

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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by polymerase View Post
    \int \frac{\sec\:x\;\cos\:2x}{\sin\;x+\sec\;x}\;dx

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    IDK if this is the best way to get the answer, but it's how I found it. Fist, cos2x = cos^2x-sin^2x so
    =\int \frac{sec[(x)cos^2(x)-sin^2(x)]}{sin(x)+sec(x)}dx

    multiply by \frac{cos(x)}{cos(x)}
    =\int \frac{cos^2(x)-sin^2(x)}{sin(x)cos(x)+1}dx

    Mess with the denominator:
    =\int \frac{cos^2(x)-sin^2(x)}{\frac 12[2sin(x)cos(x)]+1}dx


    \begin{array}{|c|}<br />
\hline And~you~should~know~that:\\<br />
\hline\\<br />
cos^2(x)=\frac 12[1+cos(2x)]\\\\<br />
sin^2(x)=\frac 12[1-cos(2x)]\\\\<br />
2sin(x)cos(x)=sin(2x)\\\\\hline\end{array}


    And now we can use these rules:
    =\int \frac{\frac 12(1+cos(2x)-\frac 12(1-cos(2x)}{\frac 12sin(2x)+1}dx

    Simplify
    =\int \frac{\frac 12+\frac 12cos(2x)-\frac 12+\frac 12cos(2x)}{\frac 12sin(2x)+1}dx

    Simplify
    =\int \frac{cos(2x)}{\frac 12sin(2x)+1}dx


    \begin{array}{|c|}<br />
\hline SUBSTITUTION\\<br />
\hline a=sin(2x)\\<br />
da =  2cos(2x)~dx\\<br />
\frac 12~da = cos(2x)~dx\\<br />
\hline\end{array}

    Make substitution
    =\frac 12\int \frac{1}{\frac 12a+1}da

    Distribute the 2 into the denominator
    =\int \frac{1}{a+2}da

    Integrate
    =ln|a+2|+C

    Anti-substitute
    =ln|sin(2x)+2|+C

    I'm pretty sure that's right, at the very least, I like all the steps I used to get there.
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  3. #3
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    Hello, polymerase!

    This is a shorter version of angel.white's solution . . .


    \int \frac{\sec x\cos2x}{\sin x+\sec x}\,dx

    Multiply by \frac{2\cos x}{2\cos x}\!:\;\;\int\frac{2\cos2x}{2\sin x\cos x + 2}\,dx \;=\;\int\frac{2\cos2x}{\sin2x + 2}\,dx

    Let u \:=\:\sin2x + 2\quad\Rightarrow\quad du \:=\:2\cos2x\,dx

    Substitute: . \int\frac{du}{u} \;=\;\ln|u| + C

    Back-substitute: . \ln|\sin2x + 2| + C

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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, polymerase!

    This is a shorter version of angel.white's solution . . .



    Multiply by \frac{2\cos x}{2\cos x}\!:\;\;\int\frac{2\cos2x}{2\sin x\cos x + 2}\,dx \;=\;\int\frac{2\cos2x}{\sin2x + 2}\,dx

    Let u \:=\:\sin2x + 2\quad\Rightarrow\quad du \:=\:2\cos2x\,dx

    Substitute: . \int\frac{du}{u} \;=\;\ln|u| + C

    Back-substitute: . \ln|\sin2x + 2| + C

    Well sure, but how are you going to scare anyone with that?!
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Fast solutions always are found with practice and of course, experience. (Smart people can get fast solutions too.)
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  6. #6
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    I just tripped over the most amazing approach . . .

    \int \frac{\sec x \cos 2x}{\sin x+\sec x}\,dx

    Multiply by \frac{\cos x}{\cos x}\!:\;\;\int \frac{\cos2x}{\sin x\cos x + 1}\,dx

    Let u \:=\:\sin x\cos x + 1

    . . Then: . du \:=\:\left[(\sin x)(-\sin x) + (\cos x)(\cos x)\right]\,dx\;=\;(\cos^2\!x-\sin^2\!x)\,dx \;=\;\cos2x\,dx

    Substitute: . \int\frac{du}{u} \;=\;\ln|u| + C

    Back-substitute: . \ln|\sec x\cos x + 1| + C . . . ta-DAA!


    Is that scary enough?

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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post

    Is that scary enough?
    Terrifying, Soroban, just terrifying!
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  8. #8
    Super Member angel.white's Avatar
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    Gonna have to check for it under my bed tonight with a flashlight.
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  9. #9
    Math Engineering Student
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    Well anyway to contemplate u' it wasn't necessary to apply the product rule, just note \sin x\cos x = \frac{{\sin 2x}}<br />
{2} and the rest follows.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Well anyway to contemplate u' it wasn't necessary to apply the product rule, just note \sin x\cos x = \frac{{\sin 2x}}<br />
{2} and the rest follows.
    yup! but that would make it less scary

    we have to be quick on one hand, but difficult on the other...
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