0-pi-4-tsint-1-cos-2-t-dt-0-pi-4-tcost-1-sin-2-t-dt-true-or-false-

Question Number 149023 by ArielVyny last updated on 02/Aug/21

\int_{0}^{\frac{π}{4}} \frac{t s i n t}{1 + {c o s}^{2} t} d t = {\int^{\frac{π}{4}}}_{0} \frac{t c o s t}{1 + {s i n}^{2} t} d t

t r u e o r f a l s e ? ?

Answered by mindispower last updated on 02/Aug/21

\frac{s i n (t)}{1 + {c o s}^{2} (t)} < \frac{c o s (t)}{1 + {s i n}^{2} (t)}, \forall t \in [0, \frac{π}{4} [

p r o f f c o s (t) > s i n (t) ⩾ 0 \Rightarrow

\frac{1}{1 + {c o s}^{2} (t)} < \frac{1}{1 + {s i n}^{2} (t)} \Rightarrow \frac{s i n (t)}{1 + {c o s}^{2} (t)} < \frac{c o s (t)}{1 + {s i n}^{2} (t)}

\Rightarrow 0 ⩽ t \frac{s i n (t)}{1 + {c o s}^{2} (t)} < \frac{t c o s (t)}{1 + {s i n}^{2} (t)}

\int_{0}^{\frac{π}{4}} \frac{t s i n (t)}{1 + {c o s}^{2} (t)} d t < \int_{0}^{\frac{π}{4}} \frac{t c o s (t)}{1 + {s i n}^{2} (t)} d t

Commented by mindispower last updated on 03/Aug/21

y e s w i t h e s p e c i a l f u n c t i o n

Commented by ArielVyny last updated on 02/Aug/21

s i r y o u c a n f i n d t h e v a l u e o f t h i s i n t e g r a l ?

Commented by ArielVyny last updated on 06/Aug/21

y o u c a n s o l v e i t p l e a s e