Evaluate the following integrals using integration By Parts 1. ∫_((π )/4) ^(π/2) xcsc^2 xdx 2. ∫


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Question Number 158839 by EbrimaDanjo last updated on 09/Nov/21

$Evaluate the following integrals using$ $integration By Parts$ $1 . \int_{\frac{π}{4}}^{\frac{π}{2}} {x c s c}^{2} x d x$ $2 . \int_{1}^{\sqrt{3}} a r c t a n (\frac{1}{x}) d x$
	Answered by gsk2684 last updated on 09/Nov/21

	$\underset{\frac{π}{4}}{\int^{\frac{π}{2}}} x {cosec}^{2} x d x$ $Missing \left or extra \right$ $Missing \left or extra \right$ $Missing \left or extra \right$ $Missing \left or extra \right$ $= [- \frac{π}{2} \cot \frac{π}{2} + \log \sin \frac{π}{2}]$ $- [- \frac{π}{4} \cot \frac{π}{4} + \log \sin \frac{π}{4}]$ $= [- \frac{π}{2} (0) + \log 1]$ $- [- \frac{π}{4} (1) + \log \frac{1}{\sqrt{2}}]$ $= [0 + 0] - [- \frac{π}{4} - \log \sqrt{2}]$ $= \frac{π}{4} + \log \sqrt{2} = \frac{π}{4} + \log 2^{\frac{1}{2}}$ $= \frac{π}{4} + \frac{1}{2} \log 2$ $. . . . . . . g s k . . . I n d i a . . . .$
	Answered by puissant last updated on 09/Nov/21
	$K=∫_1 ^(√3) arctan((1/x))dx IBP → { ((u=arctan((1/x)))),((v′=1 )) :} ⇒ { ((u′=((−(1/x^2 ))/(1+(1/x^2 ))))),((v=x)) :} ⇒ K=[xarctan((1/x))]_1 ^(√3) +∫_1 ^(√3) (x/(1+x^2 ))dx ⇒ K={(√3)arctan(((√3)/3))−(π/4)}+(1/2)[ln(1+x^2 )]_1 ^(√3) ⇒ K= (((√3)π)/6)−(π/4)+ln2 ∴∵ K = π(((√3)/6)−(1/4))+ln2.. .................Le puissant................$
	$K = \int_{1}^{\sqrt{3}} a r c t a n (\frac{1}{x}) d x$ $I B P \to {\begin{cases} u = a r c t a n (\frac{1}{x}) \\ v^{'} = 1 \end{cases} \Rightarrow {\begin{cases} u^{'} = \frac{- \frac{1}{x^{2}}}{1 + \frac{1}{x^{2}}} \\ v = x \end{cases}$ $\Rightarrow K = {[x a r c t a n (\frac{1}{x})]}_{1}^{\sqrt{3}} + \int_{1}^{\sqrt{3}} \frac{x}{1 + x^{2}} d x$ $\Rightarrow K = {\sqrt{3} a r c t a n (\frac{\sqrt{3}}{3}) - \frac{π}{4}} + \frac{1}{2} {[l n (1 + x^{2})]}_{1}^{\sqrt{3}}$ $\Rightarrow K = \frac{\sqrt{3} π}{6} - \frac{π}{4} + l n 2$ $∴∵ K = π (\frac{\sqrt{3}}{6} - \frac{1}{4}) + l n 2 . .$ $. . . . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . . . . . .$
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