prove-that-0-pi-x-tanx-tanx-secx-dx-pi-2-pi-2-

Question Number 18223 by Arnab Maiti last updated on 17/Jul/17

prove that \int_{0}^{π} \frac{x tanx}{tanx + secx} dx = \frac{π}{2} (π - 2)

Answered by alex041103 last updated on 17/Jul/17

F i r s t w e w i l l i n t e g r a t e b y p a r t s

u = x d v = \frac{t a n x}{t a n x + s e c x}

d u = d x

N o w

\frac{t a n x}{t a n x + s e c x} = \frac{t a n x (s e c x - t a n x)}{{s e c}^{2} x - {t a n}^{2} x} =

= t a n x (s e c x - t a n x) =

= s e c x t a n x - {t a n}^{2} x

S o v = s e c x - t a n x + x

N o w w e i n t e g r a t e

\int_{0}^{π} \frac{x tanx}{tanx + secx} dx =

\int_{0}^{π} x (s e c x t a n x - {t a n}^{2} x) d x =

= x {(s e c x - t a n x + x)}_{0}^{π} - \int_{0}^{π} (s e c x - t a n x + x) d x

= π (π - 1) - {[l n ∣ s e c x + t a n x ∣ + l n ∣ c o s x ∣ + \frac{x^{2}}{2}]}_{0}^{π} =

= π^{2} - π - {[l n ∣ 1 + s i n x ∣ + \frac{x^{2}}{2}]}_{0}^{π} =

= \frac{π^{2}}{2} - π = \frac{π}{2} (π - 2)

\Rightarrow \int_{0}^{π} \frac{x t a n x}{s e c x + t a n x} d x = \frac{π}{2} (π - 2)

Commented by Arnab Maiti last updated on 17/Jul/17

Thank you .

please solve your this question

\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} dx

Commented by alex041103 last updated on 17/Jul/17

i s i t Q . 17463

Commented by alex041103 last updated on 17/Jul/17

y o u c a n s e e t h e a n s w e r a n d s o l u t i o n

a t Q . 17463