calculate-0-pi-2-cos-2-x-3sin-2-x-dx-

Question Number 40823 by math khazana by abdo last updated on 28/Jul/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \sqrt{{c o s}^{2} x + 3 {s i n}^{2} x} d x

Commented by maxmathsup by imad last updated on 31/Jul/18

l e t I = \int_{0}^{\frac{π}{2}} \sqrt{{c o s}^{2} x + 3 {s i n}^{2} x} d x

I = \int_{0}^{\frac{π}{2}} c o s x \sqrt{1 + 3 {t a n}^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{\sqrt{1 + 3 {t a n}^{2} x}}{\sqrt{1 + {t a n}^{2} x}} d x c h \sqrt{3} t a n x = s h (u)

= \int_{0}^{\infty} \frac{\sqrt{1 + {s h}^{2} u}}{\sqrt{1 + \frac{{s h}^{2} u}{3}}} \frac{\frac{1}{\sqrt{3}} c h u}{(1 + \frac{{s h}^{2} u}{3})} d u (x = a r c t a n (\frac{s h u}{\sqrt{3}}))

= \frac{1}{\sqrt{3}} \int_{0}^{\infty} \frac{{c h}^{2} u d u}{\sqrt{3 + {s h}^{2} u} (3 + {s h}^{2} u)} 3 \sqrt{3} d u = 3 \int_{0}^{\infty} \frac{{c h}^{2} u}{{(3 + {s h}^{2} u)}^{\frac{3}{2}}} d u

= 3 \int_{0}^{\infty} \frac{\frac{1 + c h (2 u)}{2}}{{(3 + \frac{c h (2 u) - 1}{2})}^{\frac{3}{2}}} d u = \frac{3}{2} 2^{\frac{3}{2}} \int_{0}^{\infty} \frac{1 + c h (2 u)}{{(5 + c h (2 u))}^{\frac{3}{2}}} d u

= 3 \sqrt{2} \int_{0}^{\infty} \frac{1 + \frac{e^{2 u} + e^{- 2 u}}{2}}{{(5 + \frac{e^{2 u} + e^{- 2 u}}{2})}^{\frac{3}{2}}} d u

= 3 \sqrt{2} . \frac{2^{\frac{3}{2}}}{2} \int_{0}^{\infty} \frac{2 + e^{2 u} + e^{- 2 u}}{{(10 + e^{2 u} + e^{- 2 u})}^{\frac{3}{2}}} d u

=_{e^{u} = t} 6 \int_{1}^{+ \infty} \frac{2 + t^{2} + t^{- 2}}{{(10 + t^{2} + t^{- 2})}^{\frac{3}{2}}} \frac{d t}{t}

= 6 \int_{1}^{+ \infty} \frac{2 t^{2} + t^{4} + 1}{t^{3} {(\frac{10 t^{2} + t^{4} + 1}{t^{2}})}^{\frac{3}{2}}} d t = 6 \int_{0}^{\infty} \frac{2 t^{2} + t^{4} + 1}{{(t^{4} + 10 t^{2} + 1)}^{\frac{3}{2}}} d t \dots b e c o n t i n u e d \dots