evaluate i.∫ (((x+1)/(x−1)))dx ii. ∫_0 ^π (2cosxsinx)dx iii. ∫


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Question Number 60739 by Forkum Michael Choungong last updated on 25/May/19

$e v a l u a t e$ $i . \int (\frac{x + 1}{x - 1}) d x$ $i i . \int_{0}^{π} (2 c o s x s i n x) d x$ $i i i . \int_{\frac{π}{3}}^{π} (\frac{s i n 2 x}{c o s 2 x}) d x$
Commented by malwaan last updated on 25/May/19

$i i i . W R O N G$ $b e c a u s e \frac{3 π}{4} \in [\frac{π}{3}; π]$ $\Rightarrow 2 x = 2 \times \frac{3 π}{4} = \frac{3 π}{2}$ $\Rightarrow c o s \frac{3 π}{2} = 0$ $\Rightarrow t h e i n t e g r a l i s n o t c o n v e r g e$
Commented by maxmathsup by imad last updated on 25/May/19

$3) l e t I = \int \frac{s i n (2 x)}{c o s (2 x)} d x \Rightarrow I =_{2 x = t} \frac{1}{2} \int \frac{s i n t}{c o s t} d t$ $= - \frac{1}{2} l n ∣ c o s t ∣ + c = - \frac{1}{2} l n ∣ c o s (\frac{x}{2}) ∣ + c \Rightarrow \int_{\frac{π}{3}}^{ξ} \frac{s i n (2 x)}{c o s (2 x)} d x =$ $= {[- \frac{1}{2} l n ∣ c o s (\frac{x}{2}) ∣]}_{\frac{π}{3}}^{ξ} = \frac{1}{2} l n ∣ \frac{\sqrt{3}}{2} ∣ - \frac{1}{2} l n ∣ c o s (\frac{ξ}{2}) ∣$ ${l i m}_{ξ \to π} l n ∣ c o s (\frac{ξ}{2}) ∣ = \infty \Rightarrow t h i s i n t e g r a l d i v e r g e s . . .!$
Commented by Forkum Michael Choungong last updated on 25/May/19

$o k a y t h a n k s i d i d n t s e e t h a t$
Commented by Forkum Michael Choungong last updated on 25/May/19

$o k a y t h a n k s i d i d n t s e e t h a t$
	Answered by kaivan.ahmadi last updated on 25/May/19

	$i .$ $\int \frac{x - 1 + 2}{x - 1} d x = \int d x + \int \frac{2}{x - 1} = x + 2 l n (x - 1) + C$ $i i .$ $\int_{0}^{π} s i n 2 x d x = \frac{- 1}{2} c o s 2 x ∣_{0}^{π} = \frac{- 1}{2} (c o s 2 π - c o s 0) = 0$ $i i i .$ $u = c o s 2 x \Rightarrow d u = \frac{- 1}{2} s i n 2 x d x$ $- 2 \int \frac{d u}{u} = - 2 l n u = - 2 l n (c o s 2 x) ∣_{\frac{π}{3}}^{π} = - 2 (l n (c o s 2 π) - l n (c o s (\frac{2 π}{3})))$ $- 2 (l n 0 - l n (\frac{- 1}{2})) t h a t i s n o t t r u e, c h e k t h e q u e s t i o n$
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