∗∗∗ calculus (I) ∗∗∗ please evaluate:: φ=∫(dx/(sin(2x)ln(tan(x)))) Trinity College Cambridge ....1897...


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Question Number 131849 by mnjuly1970 last updated on 09/Feb/21

$* * * c a l c u l u s (I) * * *$ $p l e a s e e v a l u a t e ::$ $ϕ = \int \frac{d x}{s i n (2 x) l n (t a n (x))}$ $T r i n i t y C o l l e g e$ $C a m b r i d g e . . . . 1897 . . .$
	Answered by mindispower last updated on 09/Feb/21

	$\emptyset = \int \frac{d x}{2 s i n (x) c o s (x) l n (t g (x))}$ $= \int \frac{\frac{d x}{{c o s}^{2} (x)}}{2 t g (x) l n (t g (x))} l e t u = t g (x)$ $\Leftrightarrow \int \frac{d u}{2 u, l n (u)} = \frac{1}{2} \int \frac{d l n (u)}{l n (u)} = l n (l n (u)) + c$ $w e g e t l n (l n (t g (x))) + c$
	Commented by mnjuly1970 last updated on 09/Feb/21

	$t h a n k s a l o t m r m i n d s p o w e r . .$
	Answered by Dwaipayan Shikari last updated on 09/Feb/21

	$\int \frac{d x}{2 s i n x c o s x l o g (t a n x)} t a n x = t$ $= \int \frac{d t}{2 t l o g (t)} = \frac{1}{2} l o g (l o g (t)) = \frac{1}{2} l o g (l o g (t a n x)) + C$
	Commented by mnjuly1970 last updated on 09/Feb/21

	$m e r c e y m r d w a i p a y a n . . .$
	Answered by mnjuly1970 last updated on 09/Feb/21

	$ϕ = \frac{1}{2} \int \frac{t a n (x) + c o t (x)}{l n (t a n (x))} d x$ $= \frac{1}{2} \int \frac{1 + {t a n}^{2} (x)}{t a n (x) l n (t a n (x))} d x$ $\overset{t a n (x) = u}{=} \frac{1}{2} \int \frac{d u}{u l n (u)} = \frac{1}{2} \int \frac{\frac{1}{u}}{l n (u)} d u$ $= \frac{1}{2} l n (l n (t a n (x)) + C$
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