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Question Number 108748 by john santu last updated on 18/Aug/20

	Answered by bemath last updated on 19/Aug/20
	$by parts { ((u=ln (sin x) ⇒du=((cos x)/(sin x)) dx)),((v=∫ sin x dx=−cos x)) :} J=−cos x ln (sin x)+∫ ((cos^2 x dx)/(sin x)) J=−cos x ln (sin x)+∫ ((1−sin^2 x)/(sin x)) dx J=−cos x ln (sin x)+∫ cosec x dx −∫sin xdx J=−cos xln (sin x)+cos x+ln ∣cosec x−cot x∣ J={cos x (1−ln (sin x))+ln ∣((1−cos x)/(sin x))∣}_0 ^(π/2) J=lim_(b→0) {cos x(1−ln (sin x))+ln ∣((1−cos x)/(sin x))∣}_( b) ^(π/2) J= cos b (1−ln (sin b))+ln ∣((1−cos b)/(sin b))∣; where b→0 J=1$
	$b y p a r t s$ ${\begin{cases} u = \ln (\sin x) \Rightarrow d u = \frac{\cos x}{\sin x} d x \\ v = \int \sin x d x = - \cos x \end{cases}$ $J = - \cos x \ln (\sin x) + \int \frac{\cos^{2} x d x}{\sin x}$ $J = - \cos x \ln (\sin x) + \int \frac{1 - \sin^{2} x}{\sin x} d x$ $J = - \cos x \ln (\sin x) + \int cosec x d x - \int \sin x d x$ $J = - \cos x \ln (\sin x) + \cos x + \ln ∣ cosec x - \cot x ∣$ $J = {\cos x (1 - \ln (\sin x)) + \ln ∣ \frac{1 - \cos x}{\sin x} ∣}_{0}^{π / 2}$ $J = \lim_{b \to 0} {\cos x (1 - \ln (\sin x)) + \ln ∣ \frac{1 - \cos x}{\sin x} ∣}_{b}^{π / 2}$ $J = \cos b (1 - \ln (\sin b)) + \ln ∣ \frac{1 - \cos b}{\sin b} ∣; w h e r e b \to 0$ $J = 1$
	Commented by bemath last updated on 19/Aug/20

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