nice-calculus-find-0-1-sin-x-sin-1-x-dx-x-

Question Number 133107 by mnjuly1970 last updated on 18/Feb/21

\dots n i c e c a l c u l u s \dots

f i n d :: ϕ \overset{? ? ?}{=} \int_{0}^{1} (s i n (x) + s i n (\frac{1}{x})) \frac{d x}{x}

Answered by Dwaipayan Shikari last updated on 18/Feb/21

Φ = \int_{0}^{1} (s i n (x) + s i n (\frac{1}{x})) \frac{d x}{x} \frac{1}{x} = u \Rightarrow - \frac{1}{x^{2}} = \frac{d u}{d x}

= \int_{1}^{\infty} \frac{s i n (\frac{1}{u})}{u} + \frac{s i n u}{u} d u = \int_{0}^{\infty} \frac{s i n (\frac{1}{u})}{u} + \frac{s i n u}{u} d u - Φ

2 Φ = \int_{0}^{\infty} \frac{s i n u}{u} + \frac{1}{u} . s i n (\frac{1}{u}) d u

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

= \frac{π}{4} + \frac{π}{4} = \frac{π}{2}

Commented by mnjuly1970 last updated on 19/Feb/21

g r a t e f u l m r p a y a n \dots