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  1. Top | #25

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  3. Top | #26

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    $\int_{-1}^{1}\frac{e^{x}}{1+coshx}dx$

    $=\int_{-1}^{1}\frac{sihx+coshx}{1+coshx}dx$

    $=\int_{-1}^{1}\frac{sihx}{1+coshx}dx+\int_{-1}^{1}\frac{coshx}{1+coshx}dx$

    $=\int_{-1}^{1}\frac{sihx}{1+coshx}dx=0,\frac{sihx}{1+coshx } odd. function$

    $\int_{-1}^{1}\frac{coshx}{1+coshx}dx=\int_{-1}^{1}\frac{2cosh^{2}\frac{x}{2}-1}{2cosh^{2}\frac{x}{2}}dx$

    $=\int_{-1}^{1}(1-\frac{1}{2cosh^{2}\frac{x}{2}})dx=\int_{-1}^{1}(1-{\frac{1}{2}sech^{2}\frac{x}{2}})dx$

    $=x-tanh\frac{x}{2}]_{-1}^{1}=\frac{4}{e+1}$
    ; 06-20-2016 07:54 AM

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  5. Top | #27

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    $Q13)\int_{\frac{1}{\pi }}^{\frac{2}{\pi }}\frac{sin(\frac{1}{x})}{x^{2}}dx$

    $Q14)\int_{0}^{\frac{\pi }{2}}\frac{sinx}{1+sinx+cosx}dx$

    $Q15)\int_{\frac{\pi}{4}}^{\frac{\pi }{2}}\frac{cosx}{sin3x-3sinx}dx$

    $Q16)\int_{0}^{\pi }\frac{x}{1+sinx}dx$

    $Q17)\int_{0}^{\frac{\pi }{2}}\frac{x}{\sqrt{2}sin(\frac{\pi }{4}+x)}dx$

    $Q18)\int_{0}^{\frac{\pi }{2}}\frac{\sqrt{2}cos(\frac{\pi }{4}+x)}{1+sinxcosx}dx$

    $Q19)\int_{0}^{\frac{\pi }{2}}sin2x arctan(sinx)dx$

    $Q20)\int_{0}^{\pi }\frac{xcotx}{secx cscx}dx$

  6. Top | #28

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    $\large {\color{DarkBlue} Q14)\int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+sinx+cos x}dx=\int_{0}^{\pi/2}\frac{cosx}{1+sinx+cosx}dx}$
    $ {\color{DarkBlue} \Rightarrow 2I=\int_{0}^{\pi/2}(1-\frac{1}{1+sinx+cosx})dx=\frac{\pi}{2}-M}$
    ${\color{DarkBlue} M=\int_{0}^{\pi/2}\frac{1}{1+sinx+cosx}dx,,put u=tan(\frac{x}{2})}$
    ${\color{DarkBlue} M=\int_{0}^{1}\frac{\frac{2du}{1+u^{2}}}{1+\frac{1-u^{2}}{1+u^{2}}+\frac{2u}{1+u^{2}}}}$
    ${\color{DarkBlue} M=\int_{0}^{1}\frac{du}{1+u}=\ln(1+u)=\ln2}$
    ${\color{DarkBlue} 2I=\pi/2-ln2\Rightarrow I=\frac{\pi}{4}-\frac{ln2}{2}}$
    ; 06-20-2016 05:10 PM

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  8. Top | #29

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    $\large {\color{Red} Q13)\int_{1/\pi}^{2/\pi}\frac{sin(1/x)}{x^{2}}dx=\int_{1/\pi}^{2/\pi}d[-cos(1/x)]}$
    $ {\color{Red} =-[\cos(\frac{1}{x})]_{1/\pi}^{2/\pi}=-1}$
    ; 06-20-2016 06:28 PM

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  10. Top | #30

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    $\displaystyle \large {\color{Blue} \int_{\pi/4}^{\pi/2}\frac{cosx}{sin3x-3sinx}dx=\int_{\pi/4}^{\pi/2}\frac{cosx}{-4sin^{3}x}dx} $
    ${\color{Blue} =-\frac{1}{4}\int_{\pi/4}^{\pi/2}(sinx)^{-3}cosxdx=\frac{1}{8}[csc^{2}x]_{\pi/4}^{\pi/2}=-\frac{1}{8}}$

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  12. Top | #31

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  14. Top | #32

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  16. Top | #33

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  18. Top | #34

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  20. Top | #35

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    $\displaystyle \large {\color{DarkRed} I=\int_{0}^{\pi}\frac{x}{1+sinx}dx=\int_{0}^{\pi}\ frac{\pi-x}{1+sinx}dx}$
    ${\color{DarkRed} 2I=\int_{0}^{\pi}\frac{\pi}{1+sinx}dx=\pi\int_{0}^ {\pi}\frac{1}{[sin(x/2)+cos(x/2)]^{2}}dx}$
    $ {\color{DarkRed} 2I=\frac{\pi}{2}\int_{0}^{\pi}csc^{2}(\frac{x}{2}-\frac{\pi}{4})dx=-\pi[ cot(x/2-\pi/4)]_{0}^{\pi}=2\pi,,I=\pi}$
    ; 06-20-2016 06:39 PM

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  22. Top | #36

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    $Q21)\int_{0}^{\infty }\frac{e^{-2x}sinx}{x}dx$

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