calculate-0-pi-4-arctan-sinx-sinx-dx-

Question Number 87534 by mathmax by abdo last updated on 04/Apr/20

c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{a r c t a n (s i n x)}{s i n x} d x

Commented by mathmax by abdo last updated on 06/Apr/20

l e t f (a) = \int_{0}^{\frac{π}{4}} \frac{a r c t a n (a s i n x)}{s i n x} d x w i t h a > 0

f^{^{'}} (a) = \int_{0}^{\frac{π}{4}} \frac{s i n x}{(1 + a^{2} {s i n}^{2} x) s i n x} d x = \int_{0}^{\frac{π}{4}} \frac{d x}{1 + a^{2} \times \frac{1 - c o s (2 x)}{2}}

= 2 \int_{0}^{\frac{π}{4}} \frac{d x}{2 + a^{2} - a^{2} c o s (2 x)} =_{2 x = t} 2 \int_{0}^{\frac{π}{2}} \frac{d t}{2 (2 + a^{2} - a^{2} c o s t)}

=_{t a n (\frac{t}{2}) = u} \int_{0}^{1} \frac{2 d u}{(1 + u^{2}) (2 + a^{2} - a^{2} \frac{1 - u^{2}}{1 + u^{2}})}

= \int_{0}^{1} \frac{2 d u}{2 + a^{2} + (2 + a^{2}) u^{2} - a^{2} + a^{2} u^{2}} = \int_{0}^{1} \frac{2 d u}{2 + (2 + 2 a^{2}) u^{2}}

= \int_{0}^{1} \frac{d u}{1 + (1 + a^{2}) u^{2}} =_{\sqrt{1 + a^{2}} u = z} \int_{0}^{\sqrt{1 + a^{2}}} \frac{d z}{\sqrt{1 + a^{2}} (1 + z^{2})}

= \frac{1}{\sqrt{1 + a^{2}}} a r c t a n (\sqrt{1 + a^{2}}) \Rightarrow f (a) = \int_{0}^{a} \frac{a r c t a n (\sqrt{1 + α^{2}})}{\sqrt{1 + α^{2}}} d α + c

c = f (0) = 0 \Rightarrow f (a) = \int_{0}^{a} \frac{a r c t a n (\sqrt{1 + α^{2})}}{\sqrt{1 + α^{2}}} d α

a n d \int_{0}^{\frac{π}{4}} \frac{a r c t a n (s i n x)}{s i n x} d x = \int_{0}^{1} \frac{a r c t a n (\sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} d x \dots . b e c o n t i n u e d \dots