HSC integration question (1 Viewer)

sadpwner · Jul 16, 2015

I have the solution but I don't get how they got dx=2/(1+t^2) dt initially. Apparently, they differentiated t=tanx/2. How do you get that result from that.

InteGrand · Jul 16, 2015

sadpwner said:
I have the solution but I don't get how they got dx=2/(1+t^2) dt initially. Did they differentiate the t formula for cosx? I did that but I can't get that result.

$t=\tan \frac{x}{2}\Rightarrow \frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2}\sec^2 \frac{x}{2}$

$$\Rightarrow \frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2}\left( \tan^2 \frac{x}{2}+1 \right) $ (trig. identity)$

$$\Rightarrow \frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2}\left(t^2+1 \right) $ (as $t=\tan \frac{x}{2}$)$

$$\Rightarrow \frac{\mathrm{d}x}{\mathrm{d}t}=\frac{2}{1+t^2}$ (inverting)$

$$\Rightarrow \mathrm{d}x=\frac{2}{1+t^2}\text{ d}t$.$

$Alternatively, $t=\tan \frac{x}{2}\Rightarrow \frac{x}{2}=\tan^{-1}t$$

$\Rightarrow x= 2\tan^{-1}t$

$$\Rightarrow \mathrm{d}x=\frac{2}{1+t^2}\text{ d}t$.$

HSC integration question (1 Viewer)

sadpwner

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InteGrand

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