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Question Number 124587 by mnjuly1970 last updated on 04/Dec/20

\dots n i c e ◂ :::: ▸ c a l c u l u s

s i m p l e q u e s t i o n ::

p r o v e t h a t ::

\int_{0}^{\infty} \frac{4}{\sqrt{4 + x^{4}}} d x \overset{? ? ?}{=} \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{s i n (x)}} + \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{c o s (x)}}

Answered by mindispower last updated on 04/Dec/20

x = \sqrt{2} t

= \int \frac{2 \sqrt{2} d t}{\sqrt{1 + t^{4}}}

t = \sqrt{t g (x)}

d t = \frac{1}{2 {c o s}^{2} (x) \sqrt{t g (x)}}

= \frac{\sqrt{2}}{\sqrt{t g (x)}} . \frac{1}{c o s (x)} d x = \int_{0}^{\frac{π}{2}} \frac{\sqrt{2}}{\sqrt{s i n (x) c o s (x)}} d x

= \int_{0}^{\frac{π}{2}} \frac{2 d x}{\sqrt{2 s i n (x) c o s (x)}} 2 \int \frac{d x}{\sqrt{s i n (2 x)}}

2 x = w

\Rightarrow= \int_{0}^{π} \frac{d w}{\sqrt{s i n (w)}} = \int_{0}^{\frac{π}{2}} \frac{d w}{\sqrt{s i n (w)}} + \int_{\frac{π}{2}}^{π} \frac{d w}{\sqrt{s i n (w)}} ∣_{w = \frac{π}{2} + x}

= \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{s i n (x)}} + \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{c o s (x)}}

Commented by mindispower last updated on 05/Dec/20

w i t h e p l e s u r

Commented by mnjuly1970 last updated on 04/Dec/20

e x c e l l e n t . s i r m i n d s p o w e r

a s a l w a y s \dots