∫_0 ^(π/4) ((tsint)/(1+cos^2 t))dt=∫_0 ^(π/4) ((tcost)/(1+sin^2 t))dt true or false ??


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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

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