The value of
Question:

$\int \frac{d t}{\sqrt{3 t-2 t^{2}}}$

Solution:

Let, $\mathrm{I}=\int \frac{d t}{\sqrt{3 t-2 t^{2}}}=\int \frac{d t}{\sqrt{-2\left(t^{2}-\frac{3}{2} t\right)}}$

$=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-\left(t^{2}-\frac{3}{2} t+\frac{9}{16}-\frac{9}{16}\right)}} \quad$ [Making perfect

$=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-\left[\left(t-\frac{3}{4}\right)^{2}-\frac{9}{16}\right]}}=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{\frac{9}{16}-\left(t-\frac{3}{4}\right)^{2}}}$

$=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{\left(\frac{3}{4}\right)^{2}-\left(t-\frac{3}{4}\right)^{2}}}=\frac{1}{\sqrt{2}} \cdot \sin ^{-1} \frac{t-\frac{3}{4}}{\frac{3}{4}}+C$

$=\frac{1}{\sqrt{2}} \sin ^{-1} \frac{4 t-3}{3}+C$

Therefore, $\mathrm{I}=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4 t-3}{3}\right)+\mathrm{C}$.

 

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