Let f(x)= dx(3+4x^2)√4-3x^2,|x|0, then ^2+ ^2 is equal to ____________.

Question

Let $f(x)=\int \frac{dx}{(3+4{x}^{2})\sqrt{4-3{x}^{2}}},|x|0$, then ${\alpha }^{2}+{\beta }^{2}$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 28

Accepted Answer

$\begin{gathered} f(x)=\int \frac{dx}{(3+4{x}^{2})\sqrt{4-3{x}^{2}}} \ x=\frac{1}{t} \ =\int \frac{\frac{-1}{{t}^{2}}dt}{\frac{(3{t}^{2}+4)}{{t}^{2}}\frac{\sqrt{4{t}^{2}-3}}{t}} \ =\int \frac{-dt\cdot t}{(3{t}^{2}+4)\sqrt{4{t}^{2}-3}}:4{t}^{2}-3={z}^{2} \ =-\frac{1}{4}\int \frac{zdz}{(3(\frac{{z}^{2}+3}{4})+4)z} \end{gathered}$

$\begin{gathered} =\int \frac{-dz}{3{z}^{2}+25}=-\frac{1}{3}\int \frac{dz}{{z}^{2}+{(\frac{5}{\sqrt{3}})}^{2}} \ =-\frac{1}{3}\frac{\sqrt{3}}{5}{	an }^{-1}(\frac{\sqrt{3}z}{5})+C \ =-\frac{1}{5\sqrt{3}}{	an }^{-1}(\frac{\sqrt{3}}{5}\sqrt{4{t}^{2}-3})+C \ f(x)=-\frac{1}{5\sqrt{3}}{	an }^{-1}(\frac{\sqrt{3}}{5}\sqrt{\frac{4-3{x}^{2}}{{x}^{2}}})+C \ \because f(0)=0\because c=\frac{\pi }{10\sqrt{3}} \end{gathered}$

$\begin{gathered} f(1)=-\frac{1}{5\sqrt{3}}{	an }^{-1}(\frac{\sqrt{3}}{5})+\frac{\pi }{10\sqrt{3}} \ f(1)=\frac{1}{5\sqrt{3}}{\cot }^{-1}(\frac{\sqrt{3}}{5})=\frac{1}{5\sqrt{3}}{	an }^{-1}(\frac{5}{\sqrt{3}}) \ \alpha =5:\beta =\sqrt{3}	herefore {\alpha }^{2}+{\beta }^{2}=28 \end{gathered}$

Solutions · 1