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

Question Number 60893 by mathsolverby Abdo last updated on 26/May/19

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

Commented by maxmathsup by imad last updated on 27/May/19

l e t A = \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{2} {c o s}^{2} x + \sqrt{3} {s i n}^{2} x} \Rightarrow A = \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{2} \frac{1 + c o s (2 x)}{2} + \sqrt{3} \frac{1 - c o s (2 x)}{2}}

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

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

= 2 \int_{0}^{\infty} \frac{d t}{\sqrt{2} + \sqrt{3} + (\sqrt{2} + \sqrt{3}) t^{2} + \sqrt{2} - \sqrt{3}} = 2 \int_{0}^{\infty} \frac{d t}{2 \sqrt{2} + (\sqrt{2} + \sqrt{3}) t^{2}}

= \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{d t}{1 + \frac{\sqrt{2} + \sqrt{3}}{2 \sqrt{2}} t^{2}} c h a n g e m e n t \sqrt{\frac{\sqrt{2} + \sqrt{3}}{2 \sqrt{2}}} t = u g i v e

A = \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{1}{1 + u^{2}} \sqrt{\frac{2 \sqrt{2}}{\sqrt{2} + \sqrt{3}}} d u = \frac{\sqrt{\sqrt{2}}}{\sqrt{\sqrt{2} + \sqrt{3}}} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} = \frac{π}{2} \frac{\sqrt{\sqrt{2}}}{\sqrt{\sqrt{2} + \sqrt{3}}} .

Answered by Prithwish sen last updated on 27/May/19

\int \frac{dx}{\sqrt{2} \cos^{2} x + \sqrt{3} \sin^{2} x} = \int \frac{\sec^{2} x dx}{\sqrt{2} + \sqrt{3} \tan^{2} x}

Now putting tanx = v

then \sec^{2} x dx = dv

the integral changes to

\int \frac{dv}{\sqrt{2} + \sqrt{3} v^{2}}