Evaluate-the-following-integrals-using-integration-By-Parts-1-pi-4-pi-2-xcsc-2-xdx-2-1-3-arctan-1-x-dx-

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

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= [- \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

\dots \dots . g s k \dots I n d i a \dots .

Answered by puissant last updated on 09/Nov/21

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 . .

\dots \dots \dots \dots \dots . . L e p u i s s a n t \dots \dots \dots \dots \dots .

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