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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following $\int\frac{dt}{\sqrt {3t-2t^2}}$

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  • $\large\int\frac{dx}{\sqrt {a^2-x^2}}=\sin ^{-1} (\frac{x}{a})+c$
Step 1:
Let $\large\int\frac{dt}{\sqrt {3t-2t^2}}$
consider $(3t-2t^2)$
This can be written as $-(2t^2-3t)$
on completing the squares we get
$-2 \bigg[(t-\large\frac{3}{4})^2-\frac{9}{16}\bigg]$
$=2\bigg[\large\frac{9}{16}-\bigg(\frac{4t-3}{4}\bigg)^2\bigg]$
Therefore $I=\large\frac{1}{\sqrt 2} \int \Large \frac{dt}{\sqrt \frac{9}{16} -\bigg(\frac{4t-3}{4} \bigg)^2}$
Step 2:
This is of the form $\large\int\frac{dx}{\sqrt {a^2-x^2}}=\sin ^{-1} (\frac{x}{a})+c$
Here $x=\large\frac{4t-3}{4}$ and $a=\large\frac{3}{4}$
Hence $\large\frac{1}{\sqrt 2} \int \frac{dt}{\sqrt {\Large\frac {9}{16} -\bigg (\frac{4t-3}{4}\bigg)^2}}=\frac{1}{\sqrt 2} \sin ^{-1} \bigg(\Large\frac{\frac{4t-3}{4}}{\frac{3}{4}}\bigg)+c$
$=\large\frac{1}{\sqrt 2} \sin ^{-1} \frac{(4t-3)}{3}+c$
answered Apr 15, 2013 by meena.p
 
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