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Thread: another definite integral

  1. #1
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    another definite integral

    I am trying to evaluate $\displaystyle \int_{0}^{1}\frac{\tan^{-1}{x}}{x+1}dx$.

    I used integration by parts and got

    $\displaystyle \frac{\pi}{4} \ln(2) - \int_{0}^{1} \frac{\ln(x+1)}{x^2+1}dx$

    How do I solve the second integral?

    thanks!
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  2. #2
    Member sbhatnagar's Avatar
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    Lightbulb Re: another definite integral

    Let $\displaystyle I=\int_{0}^{1} \frac{\ln(x+1)}{x^2+1}dx$.

    1. Substitute $\displaystyle x = \tan{u}$ and $\displaystyle dx=\sec^2{u} \ du$.

    $\displaystyle I=\int_{0}^{\pi \over 4} \frac{\sec^2{u} \cdot \ln(\tan{u}+1)}{\sec^2{u}}dx$

    $\displaystyle \implies I=\int_{0}^{\pi \over 4} \ln(\tan{u}+1) \ dx$

    2. Use the identity $\displaystyle \int_{0}^{a}f(x) \ dx = \int_{0}^{a}f(a-x) \ dx$:

    $\displaystyle I=\int_{0}^{\pi \over 4} \ln\left(\tan{\left( \frac{\pi}{4}- u \right) }+1 \right) \ dx$

    3. Use the identity $\displaystyle \tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}$:

    $\displaystyle I=\int_{0}^{\pi \over 4} \ln\left( \frac{2}{1+\tan{u}}\right) \ dx$

    $\displaystyle \implies I=\int_{0}^{\pi \over 4} \ln(2) \ dx - \int_{0}^{\pi \over 4} \ln(1+\tan{u}) \ dx$

    $\displaystyle \implies I= \int_{0}^{\pi \over 4} \ln(2) \ dx -I$

    $\displaystyle \implies 2I=\frac{\pi}{4}\ln(2)$

    $\displaystyle \implies I=\frac{\pi}{8}\ln(2)$
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  3. #3
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    Re: another definite integral

    Quote Originally Posted by sbhatnagar View Post
    Let $\displaystyle I=\int_{0}^{1} \frac{\ln(x+1)}{x^2+1}dx$.

    1. Substitute $\displaystyle x = \tan{u}$ and $\displaystyle dx=\sec^2{u} \ du$.

    $\displaystyle I=\int_{0}^{\pi \over 4} \frac{\sec^2{u} \cdot \ln(\tan{u}+1)}{\sec^2{u}}dx$

    $\displaystyle \implies I=\int_{0}^{\pi \over 4} \ln(\tan{u}+1) \ dx$

    2. Use the identity $\displaystyle \int_{0}^{a}f(x) \ dx = \int_{0}^{a}f(a-x) \ dx$:

    $\displaystyle I=\int_{0}^{\pi \over 4} \ln\left(\tan{\left( \frac{\pi}{4}- u \right) }+1 \right) \ dx$

    3. Use the identity $\displaystyle \tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}$:

    $\displaystyle I=\int_{0}^{\pi \over 4} \ln\left( \frac{2}{1+\tan{u}}\right) \ dx$

    $\displaystyle \implies I=\int_{0}^{\pi \over 4} \ln(2) \ dx - \int_{0}^{\pi \over 4} \ln(1+\tan{u}) \ dx$

    $\displaystyle \implies I= \int_{0}^{\pi \over 4} \ln(2) \ dx -I$

    $\displaystyle \implies 2I=\frac{\pi}{4}\ln(2)$

    $\displaystyle \implies I=\frac{\pi}{8}\ln(2)$
    woa! thanks!! You used a very tricky approach.

    That should mean : $\displaystyle \int_{0}^{1}\frac{\tan^{-1}{x}}{1+x}dx=\frac{\pi}{4}\ln(2)-\frac{\pi}{8}\ln(2)=\frac{\pi}{8}\ln(2)$
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