dt-1-kt-1-t-2-0-lt-k-lt-1-

Question Number 2212 by Yozzi last updated on 09/Nov/15

\int \frac{d t}{(1 - k t) \sqrt{1 - t^{2}}} = ? 0 < k < 1

Commented by 123456 last updated on 09/Nov/15

\frac{\ln (\sqrt{k^{2} - 1} \sqrt{1 - x^{2}} + k - x) - \ln (1 - k x)}{\sqrt{k^{2} - 1}}

c . m

Commented by prakash jain last updated on 09/Nov/15

t = \sin u

d t = \cos u d u

\int \frac{d u}{1 - k \sin u}

integrate with \tan \frac{u}{2} = v

\sin u = \frac{2 v}{1 + v^{2}}

\cos u = \frac{1 - v^{2}}{1 + v^{2}}

\frac{1}{2} \sec^{2} \frac{u}{2} d u = d v

d u = \frac{2 d v}{(1 + v^{2})}

\int \frac{1}{(1 - k \frac{2 v}{1 + v^{2}})} \cdot \frac{2 d v}{(1 + v^{2})} = \int \frac{2 d v}{1 + v^{2} - 2 k v}

t o b e c o n t i n u e d i n a n s w e r .

Answered by Filup last updated on 09/Nov/15

k = 0

\int \frac{1}{(1 - 0 t) \sqrt{1 - t^{2}}} d t = \int \frac{1}{\sqrt{1 - t^{2}}}

t = \sin θ d t = \cos θ d θ

= \int \frac{\cos θ}{\sqrt{1 - \sin^{2} θ}} d θ

= \int \frac{\cos θ}{\cos θ} d θ

= \int d θ = θ + c c = constant

= \sin^{- 1} (t) + c - 1 ⩽ t ⩽ 1 (1)

k = 1

\int \frac{1}{(1 - t) \sqrt{1 - t^{2}}} d t

t = \sin θ d t = \cos θ d θ

= \int \frac{\cos θ}{(1 - \sin θ) \sqrt{1 - \sin^{2} θ}} d θ

= \int \frac{\cos θ}{(1 - \sin θ) \cos θ} d θ

= \int \frac{1}{1 - \sin θ} d θ (2)

(continue)

\sin^{- 1} (t) < \int \frac{1}{(1 - k t) \sqrt{1 - t^{2}}} d t < \int \frac{1}{1 - \sin θ} d θ

Commented by Filup last updated on 09/Nov/15

I^{'} m not sure how to finish this problem

so i hope this is partially correct and on the

right track

Answered by prakash jain last updated on 09/Nov/15

\int \frac{2 d v}{1 + v^{2} - 2 k v} = \int \frac{2 d v}{{(v - k)}^{2} + {(\sqrt{1 - k^{2}})}^{2}}

= \frac{2}{\sqrt{1 - k^{2}}} \tan^{- 1} \frac{v - k}{\sqrt{1 - k^{2}}}

v = \tan \frac{u}{2}, u = \sin^{- 1} x