Find out the answer to "Evaluating ∫03arcsin⁡(2t1+t2)dt"

Beryneingmk39

Answered question

2022-04-05

Evaluating

\int_{0}^{\sqrt{3}} \arcsin (\frac{2 t}{1 + t^{2}}) d t

Jambrichp2w2

Beginner2022-04-06Added 12 answers

We have

\frac{d}{d t} \arcsin (\frac{2 t}{1 + t^{2}}) = \frac{1}{\sqrt{1 - \frac{4 t^{2}}{{(1 + t^{2})}^{2}}}} {(\frac{2 t}{1 + t^{2}})}^{'}

= \frac{(1 + t^{2})}{\sqrt{{(1 + t^{2})}^{2} - 4 t^{2}}} (\frac{2 (1 + t^{2}) - 4 t^{2}}{{(1 + t^{2})}^{2}})

= \frac{(1 + t^{2})}{\sqrt{{(1 - t^{2})}^{2}}} (\frac{2 (1 - t^{2})}{{(1 + t^{2})}^{2}})

= \frac{2 (1 - t^{2})}{(1 + t^{2}) \sqrt{{(t^{2} - 1)}^{2}}}

= \frac{2 (1 - t^{2})}{(1 + t^{2}) | t^{2} - 1 |}

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