Let f(x)=∫dx3+4x24−3x2⋅|x| 0, then α2+β2 is equal to __

f(x)=∫dx3+4x24−3x2 Put x=1t,dx=−1t2dtf(x)=∫−dtt23+4t24−3t2=∫−tdt3t2+44t2−34t2−3=λ2⇒8tdt=2λdλ f(x)=−∫λdλ4⋅3λ2+34+4⋅λ=−∫dλ3λ2+25=−13∫dλλ2+253=−13×35tan−1⁡3λ5+cf(x)=−315tan−1⁡34−3x25x+cf(0)=0⇒c=+3π30f(1)=−315tan−1⁡35+315×π2=315cot−1⁡35 =315tan−1⁡53 =153tan−1⁡53⇒α2+β2=28

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Let $f (x) = \int \frac{dx}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}} \cdot | x | < \frac{2}{\sqrt{3}} .If f (0) = 0 and f (1) = \frac{1}{αβ} \tan^{- 1} (\frac{α}{β})$ $α, β > 0, then α^{2} + β^{2}$ is equal to __

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answer is 28.

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Detailed Solution

$\begin{array}{l} f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}} \\ Put x = \frac{1}{t}, d x = - \frac{1}{t^{2}} d t \\ f (x) = \int \frac{- d t}{t^{2} (3 + \frac{4}{t^{2}}) \sqrt{4 - \frac{3}{t^{2}}}} \\ = \int \frac{- t d t}{(3 t^{2} + 4) \sqrt{4 t^{2} - 3}} \\ 4 t^{2} - 3 = λ^{2} \Rightarrow 8 t d t = 2 λ d λ \begin{array}{l} f (x) = - \int \frac{λ d λ}{4 \cdot (3 (\frac{λ^{2} + 3}{4}) + 4) \cdot λ} \\ = - \int \frac{d λ}{3 λ^{2} + 25} = - \frac{1}{3} \int \frac{d λ}{λ^{2} + \frac{25}{3}} \\ = - \frac{1}{3} \times \frac{\sqrt{3}}{5} \tan^{- 1} (\frac{\sqrt{3} λ}{5}) + c \\ f (x) = - \frac{\sqrt{3}}{15} \tan^{- 1} (\frac{\sqrt{3} \sqrt{(4 - 3 x^{2})}}{5 x}) + c \\ f (0) = 0 \Rightarrow c = + \frac{\sqrt{3} π}{30} \end{array} \end{array}$
$\begin{array}{l} f (1) = \frac{- \sqrt{3}}{15} \tan^{- 1} (\frac{\sqrt{3}}{5}) + \frac{\sqrt{3}}{15} \times \frac{π}{2} \\ = \frac{\sqrt{3}}{15} \cot^{- 1} (\frac{\sqrt{3}}{5}) = \frac{\sqrt{3}}{15} \tan^{- 1} (\frac{5}{\sqrt{3}}) = \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{5}{\sqrt{3}}) \\ \Rightarrow α^{2} + β^{2} = 28 \end{array}$