Let I = ∫dt/√(3t 2t^2)=1/√(2)∫ dt/√( (t^2 3/2t )) =1/√(2)∫dt/√( [(t^2) 2.3/4.t+(3/4 )^2 (3/4 )^2 ]) =1/√(2)∫dt/√( [(t 3/4 )^2 (3/2 )^2]) =1/√(2)∫dt/√((3/4 ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

∫d t/√3 t-2 t...

Question

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

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Solution

$L e t I = \int \frac{d t}{\sqrt{3 t - 2 t^{2}}} = \frac{1}{\sqrt{2}} \int - \frac{d t}{\sqrt{- (t^{2} - \frac{3}{2} t)}} = \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{- [(t^{2}) - 2. \frac{3}{4} . t + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{2}]}} = \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{- [{(t - \frac{3}{4})}^{2} - {(\frac{3}{2})}^{2}]}} = \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{{(\frac{3}{4})}^{2} - {(t - \frac{3}{4})}^{2}}} = \frac{1}{\sqrt{2}} s i n^{- 1} (\frac{t - \frac{3}{4}}{\frac{3}{4}}) + C = \frac{1}{\sqrt{2}} s i n^{- 1} (\frac{4 t - 3}{3}) + C$

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∫dt√3t−2t2=

Let I=∫dt√3t−2t2=1√2∫−dt√−(t2−32t)=1√2∫dt√−[(t2)−2.34.t+(34)2−(34)2]=1√2∫dt√−[(t−34)2−(32)2]=1√2∫dt√(34)2−(t−34)2=1√2sin−1(t−3434)+C=1√2sin−1(4t−33)+C

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