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Thread: ∫(sec x tan x)/(1 + sec^2 x) dx

  1. #1
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    ∫(sec x tan x)/(1 + sec^2 x) dx

    I am asked to take this integral:

    $\displaystyle \int{\frac{\sec{x}\tan{x}}{1 + \sec^2{x}}} \: dx$

    I set $u = \sec x$, so $du = \sec{x} \tan{x} \: dx$ and the integral becomes:

    $\displaystyle \int{\frac{du}{1 + u^2}} = \arctan{u} + C = \arctan{(\sec{x})} + C$

    But the solution in the back of the book is:

    $\displaystyle -\arctan{(\cos{x})} + C$

    What am I doing wrong??
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  2. #2
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    Re: ∫(sec x tan x)/(1 + sec^2 x) dx

    nothing wrong ... the two expressions differ by a constant.

    $\displaystyle \int \dfrac{\dfrac{1}{\cos{x}} \cdot \dfrac{\sin{x}}{\cos{x}}}{1 + \dfrac{1}{\cos^2{x}}} \, dx$

    $\displaystyle \int \dfrac{\dfrac{\sin{x}}{\cos^2{x}}}{1 + \dfrac{1}{\cos^2{x}}} \, dx$

    multiply numerator & denominator by $\cos^2{x}$ ...

    $\displaystyle \int \dfrac{\sin{x}}{\cos^2{x}+1} \, dx$

    $\displaystyle -\int \dfrac{-\sin{x}}{\cos^2{x}+1} \, dx$

    $u = \cos{x} \implies du = -\sin{x} \, dx$ ...

    $\displaystyle -\int \dfrac{du}{u^2+1} = -\arctan(u) = -\arctan(\cos{x}) + C$
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  3. #3
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    Re: ∫(sec x tan x)/(1 + sec^2 x) dx

    More precisely:
    $\displaystyle \begin{aligned}y &= \arctan \sec x \\ \tan y = {\sin y \over \cos y} &= \sec y \\ \cot{\left(\tfrac\pi2 - y\right)} = {\cos{\left(\frac\pi2 - y\right)} \over \sin{\left(\frac\pi2 - y\right)}} &= \sec y \\ \tan {\left(\tfrac\pi2 - y\right)} &= \cos x \\ \tfrac\pi2 - y &= \arctan \cos x \\ y &= \tfrac\pi2 - \arctan \cos x \\[8pt] \arctan \sec x &= \tfrac\pi2 - \arctan \cos x \end{aligned}$
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