Question

Find the integrate of 2x/sqrt(1-x^2) dx from 0 to 1/2

Answer 1

Compute the definite integral:

$\large\begin{array}{l}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx}\\\\ =\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{-1}{\sqrt{1-x^2}}\cdot (-2x)\,dx\qquad\quad(i)} \end{array}$


$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{1-x^2=u~~\Rightarrow~~-2x\,dx=du}\\\\\\ \textsf{Finding the new limits of integration:}\\\\ \begin{array}{lcl} \textsf{When }\mathsf{x=0}&~\Rightarrow~&\mathsf{u=1-0^2}\\\\ &&\mathsf{u=1}\\\\\\ \textsf{When }\mathsf{x=\dfrac{1}{2}}&~\Rightarrow~&\mathsf{u=1-\left(\dfrac{1}{2}\right)^2}\\\\ &&\mathsf{u=1-\dfrac{1}{4}}\\\\ &&\mathsf{u=\dfrac{3}{4}} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{Then (i) becomes}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{\sqrt{u}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{u^{\frac{1}{2}}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}} (-1)\cdot u^{-\frac{1}{2}}\,du}\\\\ =\mathsf{(-1)\cdot \dfrac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg|_{1}^{\frac{3}{4}}} \end{array}$

$\large\begin{array}{l} =\mathsf{(-1)\cdot \dfrac{~~u^{\frac{1}{2}}~~}{\frac{1}{2}}\bigg|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{(-1)\cdot 2u^{\frac{1}{2}}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \sqrt{u}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \left(\sqrt{\dfrac{3}{4}}-\sqrt{1}\right)} \end{array}$

$\large\begin{array}{l} =\mathsf{-2\cdot \left(\dfrac{\sqrt{3}}{2}-1\right)}\\\\ =\mathsf{-\diagup\!\!\!\! 2\cdot \dfrac{\sqrt{3}}{\diagup\!\!\!\! 2}+(-2)\cdot (-1)}\\\\ =\mathsf{-\sqrt{3}+2}\\\\ =\mathsf{2-\sqrt{3}} \end{array}$


$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx=2-\sqrt{3}} \end{array}}\qquad\checkmark \end{array}$


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$\large\textsf{I hope it helps.}$


Tags: definite integral integrate limits function irrational square root sqrt fraction composite substitution integral calculus