I=∫√(1 2x)dx Substitute √(1 2x)=t⇒ 1 2x=t^2⇒ 2dx=2tdt⇒ dx=tdt I=∫ t( tdt)= ∫ t^2dt= t^33+C Undo t=√(1 2x) I= (√(1 2x))^33+C I= ( ...

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Integration by Substitution

Integrate √...

Question

Integrate $\sqrt{1 - 2 x} d x$

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Solution

$I = \int \sqrt{1 - 2 x} d x$
Substitute $\sqrt{1 - 2 x} = t \Rightarrow 1 - 2 x = t^{2} \Rightarrow - 2 d x = 2 t d t \Rightarrow - d x = t d t$
$I = \int t (- t d t) = - \int t^{2} d t = - \frac{t^{3}}{3} + C$
Undo $t = \sqrt{1 - 2 x}$
$I = - \frac{(\sqrt{1 - 2 x})^{3}}{3} + C$
$I = - \frac{(1 - 2 x)^{3 / 2}}{3} + C$

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