Substitution

**Punch** · August 21st 2011, 02:53 AM

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

**Also sprach Zarathustra** · August 21st 2011, 03:06 AM

Originally Posted by Punch

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

**NickTaken** · August 21st 2011, 03:24 AM

Originally Posted by Punch

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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