find-0-pi-x-2-cosx-sinx-dx-

Question Number 53418 by Abdo msup. last updated on 21/Jan/19

f i n d \int_{0}^{π} \frac{x}{2 + c o s x s i n x} d x

Commented by maxmathsup by imad last updated on 22/Jan/19

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

= \int_{0}^{π} \frac{2 x d x}{4 + s i n (2 x)} =_{2 x = t} \int_{0}^{2 π} \frac{t}{4 + s i n (t)} \frac{d t}{2} = \frac{1}{2} \int_{0}^{2 π} \frac{t}{4 + s i n t} d t \Rightarrow

2 I = \int_{0}^{π} \frac{t}{4 + s i n (t)} d t + \int_{π}^{2 π} \frac{t}{4 + s i n (t)} d t = H + K

K =_{t = π + u} \int_{0}^{π} \frac{π + u}{4 - s i n u} d u l e t f i n d H

H =_{t a n (\frac{t}{2}) = u} \int_{0}^{\infty} \frac{2 a r c t a n u}{4 + \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{\infty} \frac{4 a r c t a n (u)}{4 + 4 u^{2} + 2 u} d u

= 2 \int_{0}^{\infty} \frac{a r c t a n (u)}{2 u^{2} + u + 2} d u = \int_{0}^{\infty} \frac{a r c t a n (u)}{u^{2} + \frac{u}{2} + 1} d u = \int_{0}^{\infty} \frac{a r c t a n (u)}{u^{2} + 2 \frac{1}{4} u + \frac{1}{16} + 1 - \frac{1}{16}} d u

= \int_{0}^{\infty} \frac{a r c t a n (u)}{{(u + \frac{1}{4})}^{2} + \frac{15}{16}} d u =_{u + \frac{1}{4} = \frac{\sqrt{15}}{4} α} \frac{16}{15} \int_{\frac{1}{\sqrt{15}}}^{\infty} \frac{a r t a n (\frac{α \sqrt{15} - 1}{4})}{1 + α^{2}} \frac{\sqrt{15}}{4} d α

= \frac{4}{\sqrt{15}} \int_{\frac{1}{\sqrt{15}}}^{\infty} \frac{a r c t a n (\frac{α \sqrt{15} - 1}{4})}{1 + α^{2}} d α t h i s i n t e g r a l i s a t a t f o r m

\int_{λ}^{\infty} \frac{a r c t a n (a x + b)}{x^{2} + 1} d x w i c h w a n t a s p e c i a l s t u d i e s \dots

(y o u c a n c o n s i d e r f (a) = \int_{λ}^{\infty} \frac{a r c t a n (a x + b)}{x^{2} + 1} d x a n d c a l c u l a t e f i r s t f^{^{'}} (a) \dots)

w e h a v e K = π \int_{0}^{π} \frac{d u}{4 - s i n u} + \int_{0}^{π} \frac{u d u}{4 - s i n u}

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

\int_{0}^{π} \frac{d u}{4 - s i n u} =_{t a n (\frac{u}{2}) = t} \int_{0}^{\infty} \frac{1}{4 - \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int_{0}^{\infty} \frac{2 d t}{4 + 4 t^{2} - 2 t}

= \int_{0}^{\infty} \frac{d t}{2 t^{2} - t + 2} d t = \frac{1}{2} \int_{0}^{\infty} \frac{d t}{t^{2} - \frac{t}{2} + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{d t}{t^{2} - 2 \frac{1}{4} t + \frac{1}{16} + 1 - \frac{1}{16}}

= \frac{1}{2} \int_{0}^{\infty} \frac{d t}{{(t - \frac{1}{4})}^{2} + \frac{15}{16}} =_{t - \frac{1}{4} = \frac{\sqrt{15}}{4} u} \frac{1}{2} \frac{16}{15} \int_{- \frac{1}{\sqrt{15}}}^{+ \infty} \frac{1}{1 + u^{2}} \frac{\sqrt{15}}{4} d u

= \frac{2}{\sqrt{15}} {[a r c t a n (u)]}_{- \frac{1}{\sqrt{15}}}^{+ \infty} = \frac{2}{\sqrt{15}} {\frac{π}{2} + a r c t a n (\frac{1}{\sqrt{15}})}