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Math Help - Definite integrals

  1. #1
    Member helloying's Avatar
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    Definite integrals

    \frac{d}{dx}\tan x=\sec^2 x
    \frac{d}{dx}\sec x = \sec x\sin x

    Hence evaluate \int_0^\pi \frac{1 + \sin x}{\cos^2 x} dx
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  2. #2
    MHF Contributor Amer's Avatar
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    write it like that it easy now

    \int\frac{1}{cos^2x}dx+\int\frac{sinx}{cos^2x}dx

    \int sec^2xdx +\int\frac{sinx}{cos^2x}dx
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  3. #3
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    \int_0^\pi \frac{1 + \sin (x)}{\cos^2 (x)} dx


    \int_0^\pi \frac{1}{\cos^2 (x)} +\frac{ \sin (x)}{\cos^2 (x)}dx

    \int_0^\pi sec^2 (x) +\frac{ \tan (x)}{\cos(x)}dx

    \int_0^\pi sec^2 (x) + tan (x)sec(x)dx
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  4. #4
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    Quote Originally Posted by Amer View Post
    write it like that it easy now

    \int\frac{1}{cos^2x}dx+\int\frac{sinx}{cos^2x}dx

    \int sec^2xdx +\int{sinx}{cos^2x}dx

     <br />
 \int\frac{sinx}{cos^2x}dx \neq \int{sinx}{cos^2x}dx<br />
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  5. #5
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    \frac{d}{dx}\tan (x)=\sec^2( x) \Rightarrow \tan (x)=\int \sec^2( x)dx


    \frac{d}{dx}\sec( x) = \sec( x)\sin (x) \Rightarrow \sec( x) =\int \sec( x)\sin (x)dx
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