Using t-formulae in integration (1 Viewer)

ashokkumar · May 26, 2010

Using the substitution t = tan(x/2), evaluate the integral dx /(13 + 5sinx + 12cosx) from x=0 to x=pi/3.

jet · May 27, 2010

$\text{Obviously, we let } t = \tan \left (\frac{x}{2} \right ) \Rightarrow \frac{dt}{dx} = \frac{1}{2}\sec^2 \left (\frac{x}{2} \right ) = \frac{1}{2}\left (1 + \tan^2 \left (\frac{x}{2} \right ) \right ) = \frac{1 + t^2}{2} \\ \therefore dx = \frac{2\, dt}{1 + t^2} \\ x = 0, t = 0 \\ x = \frac{\pi}{3}, t = \frac{1}{\sqrt{3}} \\ \therefore\int ^{\frac{\pi}{3}}_{0} \frac{dx}{13 + 5\sin x + 12\cos x} = \int^{\frac{1}{\sqrt{3}}}_{0} \frac{\frac{2\, dt}{1 + t^2}}{13 + \frac{5\cdot 2t}{1 + t^2} + \frac{12 \left (1 - t^2 \right )}{1 + t^2}} \\ = \int^{\frac{1}{\sqrt{3}}}_0 \frac{\frac{2\, dt}{1 + t^2}}{\frac{13\left (1 + t^2 \right ) + 10t + 12 - 12t^2}{1 + t^2}} \\ = \int^{\frac{1}{\sqrt{3}}}_0 \frac{2 \, dt}{t^2 + 10t + 25} \\ = \int^{\frac{1}{\sqrt{3}}}_0 \frac{2 \, dt}{\left (t + 5 \right )^2} \\ =2 \left [\frac{-1}{t + 5} \right ]^{\frac{1}{\sqrt{3}}}_0 \\ = 2\left [\frac{-\sqrt{3}}{1 + 5\sqrt{3}} + \frac{1}{5} \right ] \\$
You should be able to simplify that further; just year 10/11 surds work.

ashokkumar · May 27, 2010

thanks

Using t-formulae in integration (1 Viewer)

ashokkumar

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jet

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ashokkumar

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