calculate-0-1-2-1-x-4-dx-

Question Number 51987 by maxmathsup by imad last updated on 01/Jan/19

c a l c u l a t e \int_{0}^{\frac{1}{2}} \sqrt{1 - x^{4}} d x

Commented by Abdo msup. last updated on 02/Jan/19

l e t A = \int_{0}^{\frac{1}{2}} \sqrt{1 - x^{4}} d x \Rightarrow A = \int_{0}^{\frac{1}{2}} \sqrt{1 - x^{2}} \sqrt{1 + x^{2}} d x

=_{x^{2} = c o s t} \int_{1}^{\frac{π}{3}} \sqrt{1 - c o s t} \sqrt{1 + c o s t} \frac{- s i n t}{2 \sqrt{c o s t}} d t

= - \int_{1}^{\frac{π}{3}} \sqrt{2} s i n (\frac{t}{2}) \sqrt{2} c o s (\frac{t}{2}) \frac{s i n t}{2 \sqrt{c o s t}} d t

= - \int_{1}^{\frac{π}{3}} s i n (\frac{t}{2}) c o s (\frac{t}{2}) \frac{s i n t}{\sqrt{c o s t}} d t

= - \int_{1}^{\frac{π}{3}} \frac{1}{2} \frac{{s i n}^{2} t}{\sqrt{c o s t}} d t = - \frac{1}{2} \int_{1}^{\frac{π}{3}} \frac{1 - {c o s}^{2} t}{\sqrt{c o s t}} d t

= - \frac{1}{2} \int_{1}^{\frac{π}{3}} \frac{d t}{\sqrt{c o s t}} + \int_{1}^{\frac{π}{3}} \frac{{c o s}^{2} t}{\sqrt{c o s t}} d t \dots .

b e c o n t i n u e d \dots

Answered by peter frank last updated on 01/Jan/19

\int \sqrt{(1 + x^{2}) (1 - x^{2})}

x^{2} = \sin θ \Rightarrow 2 x d x = \cos θ d θ

\int \sqrt{(1 + \sin θ) (1 - \sin θ)} d x

\int \sqrt{1 - \sin^{2}} \frac{\cos θ d θ}{2 x}

\frac{1}{2} \int \frac{\cos^{2} θ d θ}{x}

\frac{1}{2} \int \frac{\cos^{2} θ d θ}{\sqrt{\sin θ}}

\dots . .