QJEEM23Mabff5: JEE Main 2023 Mathematics - Class 12 JEEM23Mabff5 Solution: Let , then is equal to ____________.

Question

Verified Solution • 2025–26 Session moderate

Let

f(x)=\int \frac{d x}{\left(3+4 x^{2}\right) \sqrt{4-3 x^{2}}},|x|&lt;\frac{2}{\sqrt{3}}

. If

f(0)=0

and

f(1)=\frac{1}{\alpha \beta} \tan ^{-1}\left(\frac{\alpha}{\beta}\right)

,

\alpha, \beta&gt;0

, then

\alpha^{2}+\beta^{2}

is equal to ____________.

Accepted Answer

\begin{aligned}&amp; f(x)=\int \frac{d x}{\left(3+4 x^2\right) \sqrt{4-3 x^2}} \\\\&amp; x=\frac{1}{t} \\\\&amp; =\int \frac{\frac{-1}{t^2} d t}{\frac{\left(3 t^2+4\right)}{t^2} \frac{\sqrt{4 t^2-3}}{t}} \\\\&amp; =\int \frac{-d t \cdot t}{\left(3 t^2+4\right) \sqrt{4 t^2-3}}: \text { Put } 4 t^2-3=z^2 \\\\&amp; =-\frac{1}{4} \int \frac{z d z}{\left(3\left(\frac{z^2+3}{4}\right)+4\right) z}\end{aligned}

\begin{aligned}&amp; =\int \frac{-d z}{3 z^2+25}=-\frac{1}{3} \int \frac{\mathrm{dz}}{\mathrm{z}^2+\left(\frac{5}{\sqrt{3}}\right)^2} \\\\&amp; =-\frac{1}{3} \frac{\sqrt{3}}{5} \tan ^{-1}\left(\frac{\sqrt{3} z}{5}\right)+C \\\\&amp; =-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{4 t^2-3}\right)+C \\\\&amp; f(x)=-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{\frac{4-3 \mathrm{x}^2}{x^2}}\right)+C \\\\&amp; \because \mathrm{f}(0)=0 \because \mathrm{c}=\frac{\pi}{10 \sqrt{3}}\end{aligned}

\begin{aligned}&amp; f(1)=-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5}\right)+\frac{\pi}{10 \sqrt{3}} \\\\&amp; f(1)=\frac{1}{5 \sqrt{3}} \cot ^{-1}\left(\frac{\sqrt{3}}{5}\right)=\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{5}{\sqrt{3}}\right) \\\\&amp; \alpha=5: \beta=\sqrt{3} \therefore \alpha^2+\beta^2=28\end{aligned}

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