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Math Help - Substitution

  1. #1
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    Substitution

    Using the substitution t=tan(\frac{\theta}{2}), show that \int^{\frac{\pi}{2}}_0\frac{1}{5cos\theta+3} d{\theta} =\frac{1}{4}ln3
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Substitution

    Quote Originally Posted by Punch View Post
    Using the substitution t=tan(\frac{\theta}{2}), show that \int^{\frac{\pi}{2}}_0\frac{1}{5cos\theta+3} d{\theta} =\frac{1}{4}ln3
    \int\frac{1}{5cos(x&#41 ;+3} dx - Wolfram|Alpha
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  3. #3
    Newbie NickTaken's Avatar
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    Re: Substitution

    Quote Originally Posted by Punch View Post
    Using the substitution t=tan(\frac{\theta}{2}), show that \int^{\frac{\pi}{2}}_0\frac{1}{5cos\theta+3} d{\theta} =\frac{1}{4}ln3
    \begin{aligned}I & = \int_{0}^{\pi/2}\frac{1}{5\cos{\theta}+3}}\;{d\theta} = \int_{0}^{\pi/2}\frac{\sec{\frac{\theta}{2}}}{8-2\tan^2{\frac{\theta}{2}}}\;{d\theta} \\ & = 2\int_{0}^{1}\frac{1}{8-2t^2}\;{dt} = \int_{0}^{1}\frac{1}{\left(2\sqrt{2}\right)^2-\left(\sqrt{2}\right)^2t^2}\;{dt} \\& = \frac{1}{2\sqrt{2}\sqrt{2}}\ln\bigg|\frac{2\sqrt{2  }+\sqrt{2t}}{2\sqrt{2}-\sqrt{2}t}\bigg|_{ t=0}^{1} = \frac{1}{4}\ln(3). \end{aligned}
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