1. ## Substitution

Using the substitution $\displaystyle t=tan(\frac{\theta}{2})$, show that $\displaystyle \int^{\frac{\pi}{2}}_0\frac{1}{5cos\theta+3} d{\theta} =\frac{1}{4}ln3$

2. ## Re: Substitution

Originally Posted by Punch
Using the substitution $\displaystyle t=tan(\frac{\theta}{2})$, show that $\displaystyle \int^{\frac{\pi}{2}}_0\frac{1}{5cos\theta+3} d{\theta} =\frac{1}{4}ln3$
&#92;int&#92;frac&#123;1&#125;&#123;5cos&#40;x&#41 ;&#43;3&#125; dx - Wolfram|Alpha

3. ## Re: Substitution

Originally Posted by Punch
Using the substitution $\displaystyle t=tan(\frac{\theta}{2})$, show that $\displaystyle \int^{\frac{\pi}{2}}_0\frac{1}{5cos\theta+3} d{\theta} =\frac{1}{4}ln3$
\displaystyle \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}