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

Question Number 42804 by maxmathsup by imad last updated on 02/Sep/18

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

Commented by maxmathsup by imad last updated on 03/Sep/18

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

I =_{2 x = u} \int_{1}^{2} \frac{d u}{2 {\sqrt{u^{2} - 1} + \sqrt{u^{2} + 1}}} \Rightarrow 2 I = \int_{1}^{2} \frac{\sqrt{u^{2} + 1} - \sqrt{u^{2} - 1}}{2} d u \Rightarrow

4 I = \int_{1}^{2} \sqrt{u^{2} + 1} d u - \int_{1}^{2} \sqrt{u^{2} - 1} d u b u t

\int_{1}^{2} \sqrt{1 + u^{2}} d u =_{u = s h (x)} \int_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})} c h (x) c h (x) d x = \int_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})} \frac{1 + c h (2 x)}{2} d x

= \frac{1}{2} {l n (2 + \sqrt{5}) - l n (1 + \sqrt{2})} + \frac{1}{4} {[s h (2 x)]}_{l n (1 + \sqrt{2})}^{l n - 2 + \sqrt{5})}

= \frac{1}{2} {l n (2 + \sqrt{5}) - l n (1 + \sqrt{2})} + \frac{1}{4} {[\frac{e^{2 x} - e^{- 2 x}}{2}]}_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})}

= \frac{1}{2} {l n (2 + \sqrt{5}) - l n (1 + \sqrt{2})} + \frac{1}{8} {{(2 + \sqrt{5})}^{2} - {(2 + \sqrt{5})}^{- 2} - {(1 + \sqrt{2})}^{2} + {(1 + \sqrt{2})}^{- 2}}

a l s o w e h a v e \int_{1}^{2} \sqrt{u^{2} - 1} d u =_{u = c h (x)} \int_{0}^{l n (2 + \sqrt{3})} s h (x) s h (x + d x

= \frac{1}{2} \int_{0}^{l n (2 + \sqrt{3})} (c h (2 x) - 1) d x = - \frac{1}{2} l n (2 + \sqrt{3}) + \frac{1}{4} {[s h (2 x)]}_{0}^{l n (2 + \sqrt{3})}

= - \frac{1}{2} l n (2 + \sqrt{3}) + \frac{1}{8} {[e^{2 x} - e^{- 2 x}]}_{0}^{l n (2 + \sqrt{3})}

= - \frac{1}{2} l n (2 + \sqrt{3}) + \frac{1}{8} {{(2 + \sqrt{3})}^{2} - {(2 + \sqrt{3})}^{- 2}} s o t h e v a l u e o f I i s d e t e r m i n e d .

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18

= \frac{1}{2} \int_{\frac{1}{2}}^{1} \frac{(4 x^{2} + 1) - (4 x^{2} - 1)}{\sqrt{4 x^{2} + 1} + \sqrt{4 x^{2} - 1}} d x

= \frac{1}{2} \int_{\frac{1}{2}}^{1} \sqrt{4 x^{2} + 1} - \sqrt{4 x^{2} - 1} d x

= \int_{\frac{1}{2}}^{1} \sqrt{x^{2} + \frac{1}{2^{2}}} - \sqrt{x^{2} - \frac{1}{2^{2}}} d x

n o w u s e f o r m u l a \int \sqrt{x^{2} + a^{2}} a n d \int \sqrt{x^{2} - a^{2}} d x

∣ \frac{x \sqrt{x^{2} + \frac{1}{4}}}{2} + \frac{\frac{1}{2^{2}}}{2} l n (x + \sqrt{x^{2} + \frac{1}{4}}) - \frac{x \sqrt{x^{2} - \frac{1}{4}}}{2} + \frac{\frac{1}{2^{2}}}{2} l n (x + \sqrt{x^{2} - \frac{1}{4}}) ∣_{\frac{1}{2}}^{1}