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Thread: integration

  1. #1
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    integration

    integarte {secx}\{1+2
    sin^2(x)},dx
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: integration

    Maybe you can try with the t-formulas, let $\displaystyle t=\tan\left(\frac{x}{2}\right)$
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  3. #3
    Super Member TheChaz's Avatar
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    Re: integration

    Quote Originally Posted by ayushdadhwal View Post
    integarte {secx}\{1+2
    sin^2(x)},dx
    (Quoted from mymathforum, where you cross-posted...)
    We are given:

    $\displaystyle \int\frac{\sec x}{1+2\sin^2x}\,dx$

    Let's rewrite the integrand:

    $\displaystyle \frac{\sec x}{1+2\sin^2x}=\frac{\cos x}{3}\(\frac{1+2\sin^2x+2\(1-\sin^2x\)}{\cos^2x\(1+2\sin^2x\)}\)=\frac{\cos x}{3}\(\frac{1}{1-\sin^2x}+\frac{2}{1+2\sin^2x}\)=$

    $\displaystyle \frac{\cos x}{6}\(\frac{2}{1-\sin^2x}+\frac{4}{1+2\sin^2x}\)=\frac{\cos x}{6}\(\frac{\(\sin x-1\)-\(\sin x+1\)}{\sin^2x-1}+\frac{4}{1+2\sin^2x}\)=$

    $\displaystyle \frac{1}{6}\(\frac{\cos x}{\sin x+1}-\frac{\cos x}{\sin x-1}+2\sqrt{2}\frac{\sqrt{2}\cos x}{\(\sqrt{2}\sin x\)^2+1}\)$

    Now we may state:

    $\displaystyle \int\frac{\sec x}{1+2\sin^2x}\,dx=\frac{1}{6}\(\ln|\sin x+1|-\ln|\sin x-1|+2\sqrt{2}\tan^{\small{-1}}\(\sqrt{2}\sin x\)\)+C$

    That was MarkFL's answer...

    Almost simultaneously, galactus wrote:

    $\displaystyle \int\frac{sec(x)}{1+2sin^{2}(x)}dx$

    $\displaystyle \int\frac{sec(x)}{1+(\sqrt{2}sin(x))^{2}}dx$

    Let $\displaystyle t=\sqrt{2}sin(x), \;\ dt=\sqrt{2}cos(x)dx$

    Notice now we have the form $\displaystyle \frac{1}{1+t^{2}}$

    Continuing:

    Since $\displaystyle t=\sqrt{2}sin(x)$, then $\displaystyle sec(x)=\sqrt{\frac{2}{2-t^{2}}}$ (this just comes from looking at a right triangle).

    $\displaystyle dx=\frac{dt}{\sqrt{2}}sec(x)\Rightarrow dx=\frac{1}{\sqrt{2}}\sqrt{\frac{2}{2-t^{2}}}dt=\frac{1}{\sqrt{2-t^{2}}}dt$

    Put all together:

    $\displaystyle \int\frac{\sqrt{\frac{2}{2-t^{2}}}}{1+t^{2}}\cdot\frac{1}{2-t^{2}}dt=\frac{\sqrt{2}}{3}\int\frac{1}{t^{2}+1}dt +\frac{\sqrt{2}}{3}\int\frac{1}{2-t^{2}}dt$

    The left most integral is an arctan. The right one is of the form $\displaystyle \int\frac{1}{a^{2}-t^{2}}dt=\frac{1}{2a}ln\left|\frac{t+a}{t-a}\right|$

    So, we get after resubbing:

    $\displaystyle \frac{ln|sin(x)+1|}{6}-\frac{ln|sin(x)-1|}{6}+\frac{\sqrt{2}}{3}tan^{-1}(\sqrt{2}sin(x))$


    http://www.mymathforum.com/viewtopic...start=0#p89474
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  4. #4
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    Re: integration

    $\displaystyle \int\frac{\sec(x)}{1+2\sin^2(x)}dx=\int\frac{\cos( x)}{\cos^2(x)(1+2\sin^2(x))}dx$
    $\displaystyle =\int\frac{\cos(x)}{(1-\sin^2(x))(1+2\sin^2(x))}dx$
    substitute $\displaystyle \sin(x)=t$ to get
    $\displaystyle \int\frac{1}{(1-t^2)(1+2t^2)}dx$
    now partial fractions can be used to integrate
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  5. #5
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    Re: integration

    $\displaystyle \displaystyle \begin{align*}\int{\frac{\sec{x}}{1 + 2\sin^2{x}}\,dx} &= \int{\frac{1}{\cos{x}(1 + 2\sin^2{x})}\,dx} \\ &= \int{\frac{\cos{x}}{\cos^2{x}(1 + 2\sin^2{x})}\,dx} \\ &= \int{\frac{\cos{x}}{(1 - \sin^2{x})(1+ 2\sin^2{x})}\,dx} \\ &= \int{\frac{1}{(1 - u^2)(1 + 2u^2)}\,du} \end{align*}$

    You can now integrate this using partial fractions.
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