1-lim-x-a-x-a-cosec-pix-a-2-x-2-1-dx-by-using-Euler-s-substitution-

Question Number 112651 by bemath last updated on 09/Sep/20

(1) \lim_{x \to a} (x - a) cosec (\frac{π x}{a}) ?

(2) \int \sqrt{x^{2} + 1} dx, by using {Euler}^{'} s

substitution

Answered by john santu last updated on 09/Sep/20

(★) b y {E u l e r}^{'} s s u b s t i t u t i o n

l e t \sqrt{x^{2} + 1} = x + q \Rightarrow x^{2} + 1 = x^{2} + 2 q x + q^{2}

2 q x + q^{2} = 1 \Rightarrow x = \frac{1 - q^{2}}{2 q}

d x = \frac{- 4 q^{2} - 2 (1 - q^{2})}{4 q^{2}} d q

d x = \frac{- 2 q^{2} - 2}{4 q^{2}} d q = \frac{- q^{2} - 1}{2 q^{2}} d q

s o I = \int \sqrt{x^{2} + 1} d x

I = \int (\frac{1 - q^{2}}{2 q} + q) . (\frac{- q^{2} - 1}{2 q^{2}}) d q

I = - \int (\frac{q^{2} + 1}{2 q}) (\frac{q^{2} + 1}{2 q^{2}}) d q

I = - \frac{1}{4} \int (\frac{q^{4} + 2 q^{2} + 1}{q^{3}}) d q

I = - \frac{1}{4} \int (q + \frac{2}{q} + q^{- 3}) d q

I = - \frac{1}{4} (\frac{1}{2} q^{2} + 2 \ln (q) - \frac{1}{2 q^{2}}) + c

I = - \frac{1}{8} {(\sqrt{x^{2} + 1} - x)}^{2} - \frac{1}{2} \ln ∣ \sqrt{x^{2} + 1} - x ∣ + \frac{1}{8 {(\sqrt{x^{2} + 1} - x)}^{2}} + c

Answered by bobhans last updated on 09/Sep/20

(⧫) \lim_{x \to a} (x - a) cosec (\frac{π x}{a}) = L

L = \lim_{x \to a} \frac{(x - a)}{\sin (\frac{π x}{a})} = \lim_{x \to a} \frac{1}{\frac{π}{a} \cos (\frac{π x}{a})}

L = \frac{a}{π} \lim_{x \to a} \frac{1}{\cos (\frac{π x}{a})} = - \frac{a}{π}

Answered by 1549442205PVT last updated on 10/Sep/20

2) Find \int \sqrt{x^{2} + 1} dx

Put x = tant \Rightarrow dx = (1 + \tan^{2} t) dt . Then

F = \int \sqrt{x^{2} + 1} dx = \int \sqrt{1 + \tan^{2} t} (1 + \tan^{2} t) dt

F = \int \frac{dt}{\cos^{3} t} dt = \frac{1}{2} \int \frac{1}{sint} \times \frac{2 sintcost}{\cos^{4} t} dt

= \frac{1}{2} \int \frac{1}{sint} d (\frac{1}{\cos^{2} t}) = \frac{1}{2} \times \frac{1}{{sintcos}^{2} t}

- \frac{1}{2} \int \frac{1}{\cos^{2} t} . {(\frac{1}{sint})}^{'} dt = \frac{1}{2} \times \frac{1}{{sintcos}^{2} t}

- \frac{1}{2} \int \frac{1}{\cos^{2} t} . (- \frac{1}{\sin^{2} t}) \times costdt

= \frac{1}{2} \times \frac{1}{{sintcos}^{2} t} + \frac{1}{2} \int \frac{d (sint)}{\cos^{2} {tsin}^{2} t}

Put sint = u we get

\frac{1}{2} \int \frac{d (sint)}{\cos^{2} {tsin}^{2} t} = \frac{1}{2} \int \frac{du}{u^{2} (1 - u^{2})}

= \frac{1}{2} \int (\frac{1}{u^{2}} + \frac{1}{1 - u^{2}}) du = - \frac{1}{2 u} - \frac{1}{2} \int \frac{du}{u^{2} - 1}

= \frac{- 1}{2 u} - \frac{1}{2} \times \frac{1}{2} \ln ∣ \frac{u - 1}{u + 1} ∣ . Therefore,

F = \frac{1}{2 {sintcos}^{2} t} - \frac{1}{2 sint} - \frac{1}{4} \ln ∣ \frac{sint - 1}{sint + 1} ∣ + C

x = tant \Leftrightarrow \cos^{2} t = \frac{1}{1 + x^{2}}, sint = \frac{∣ x ∣}{\sqrt{1 + x^{2}}}

F = \frac{∣ x ∣ \sqrt{1 + x^{2}}}{2} - \frac{1}{4} \ln ∣ \frac{∣ x ∣ - \sqrt{1 + x^{2}}}{∣ x ∣ + \sqrt{1 + x^{2}}} ∣ + C

Also see question 112313