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Question Number 162189 by SANOGO last updated on 27/Dec/21

s h o w t h e {c o n v e r g e}^{} n c e a n d c a l c u l a t e

\int_{- 1}^{1} \sqrt{\frac{1 - t}{1 + t}} d t

Commented by Ar Brandon last updated on 27/Dec/21

I = \int_{- 1}^{1} \sqrt{\frac{1 - t}{1 + t}} d t = \int_{- 1}^{1} \frac{1 - t}{\sqrt{1 - t^{2}}} d t

= \int_{- 1}^{1} \frac{d t}{\sqrt{1 - t^{2}}} - \int \frac{t}{\sqrt{1 - t^{2}}} d t

= {[\arcsin (t)]}_{- 1}^{1} - {[\sqrt{1 - t^{2}}]}_{- 1}^{1} = π

Commented by SANOGO last updated on 28/Dec/21

t h a n k y o u

Answered by mathmax by abdo last updated on 27/Dec/21

I = \int_{- 1}^{1} \sqrt{\frac{1 - t}{1 + t}} dt \Rightarrow I = \int_{- 1}^{0} \sqrt{\frac{1 - t}{1 + t}} dt + \int_{0}^{1} \sqrt{\frac{1 - t}{1 + t}} dt

\int_{- 1}^{0} \sqrt{\frac{1 - t}{1 + t}} dt =_{t = - u} - \int_{0}^{1} \sqrt{\frac{1 + u}{1 - u}} du

=_{u = \cos θ} - \int_{\frac{π}{2}}^{0} \sqrt{\frac{{2 \cos}^{2} (\frac{θ}{2})}{{2 \sin}^{2} (\frac{θ}{2})}} (- \sin θ) d θ

= - \int_{0}^{\frac{π}{2}} \frac{\cos (\frac{θ}{2})}{\sin (\frac{θ}{2})} \times 2 \sin (\frac{θ}{2}) \cos (\frac{θ}{2}) d θ

= - 2 \int_{0}^{\frac{π}{2}} \cos^{2} (\frac{θ}{2}) d θ = - \int_{0}^{\frac{π}{2}} (1 + \cos θ) d θ

= - \frac{π}{2} - {[\sin θ]}_{0}^{\frac{π}{2}} = - \frac{π}{2} - 1

\int_{0}^{1} \sqrt{\frac{1 - t}{1 + t}} dt =_{t = \cos θ} \int_{\frac{π}{2}}^{0} \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} (- \sin θ) d θ

= \int_{0}^{\frac{π}{2}} \sqrt{\frac{{2 \sin}^{2} (\frac{θ}{2})}{{2 \cos}^{2} (\frac{θ}{2})}} \sin θ d θ

= \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{θ}{2})}{\cos (\frac{θ}{2})} \times 2 \sin (\frac{θ}{2}) \cos (\frac{θ}{2}) d θ

= 2 \int_{0}^{\frac{π}{2}} \sin^{2} (\frac{θ}{2}) d θ = \int_{0}^{\frac{π}{2}} (1 - \cos (θ)) d θ

= \frac{π}{2} - {[\sin θ]}_{0}^{\frac{π}{2}} = \frac{π}{2} - 1 \Rightarrow

I = - \frac{π}{2} - 1 + \frac{π}{2} - 1 \Rightarrow I = - 2

Commented by SANOGO last updated on 27/Dec/21

t h a n k y o u

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