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Question Number 54074 by rahul 19 last updated on 28/Jan/19

E v a l u a t e :

1) \int_{0}^{1} \frac{d x}{\sqrt{1 + x} + \sqrt{1 - x} + 2}

2) \int_{0}^{2} \frac{l n (1 + 2 x)}{1 + x^{2}}

3)

\int_{0}^{π} \frac{x}{\sqrt{1 + \sin^{3} x}} ((3 π \cos x + 4 \sin x) \sin^{2} x + 4) d x

4)

\int_{0}^{π} \frac{x^{2} \cos^{2} x - x \sin x - \cos x - 1}{{(1 + x \sin x)}^{2}} d x .

Commented by maxmathsup by imad last updated on 28/Jan/19

1) l e t I = \int_{0}^{1} \frac{d x}{\sqrt{1 + x} + \sqrt{1 - x} + 2} c h a n g e m e n t x = c o s θ g i v e

I = - \int_{0}^{\frac{π}{2}} \frac{- s i n θ d θ}{\sqrt{2} c o s (\frac{θ}{2}) + \sqrt{2} s i n (\frac{θ}{2}) + 2} = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{s i n θ}{c o s (\frac{θ}{2}) + s i n (\frac{θ}{2}) + \sqrt{2}} d θ

=_{\frac{θ}{2} = t} \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{4}} \frac{s i n (2 t)}{c o s t + s i n t + \sqrt{2}} (2) d t = \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{s i n (2 t)}{c o s t + s i n t + \sqrt{2}} d t

= 2 \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{s i n t c o s t}{c o s t + s i n t + \sqrt{2}} d t =_{t a n (\frac{t}{2}) = x} 2 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{\frac{2 x (1 - x^{2})}{{(1 + x^{2})}^{2}}}{\frac{1 - x^{2}}{1 + x^{2}} + \frac{2 x}{1 + x^{2}} + \sqrt{2}} \frac{2 d x}{1 + x^{2}}

= 8 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{x (1 - x^{2})}{{(1 + x^{2})}^{2} (1 - x^{2} + 2 x + \sqrt{2} (1 + x^{2}))} d x

= 8 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{x (1 - x^{2})}{{(1 + x^{2})}^{2} ((\sqrt{2} - 1) x^{2} + 2 x + 1 + \sqrt{2})} d x l e t d e c o m p o s e

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

r o o t s o f (\sqrt{2} - 1) x^{2} + 2 x + 1 + \sqrt{2}

Δ = 4 - 4 (\sqrt{2} - 1) (\sqrt{2} + 1) = 0 \Rightarrow o n e r o o t x_{0} = - \frac{b}{2 a} = - \frac{2}{2 (\sqrt{2} - 1)} = \frac{- 1}{\sqrt{2} - 1} \Rightarrow

(\sqrt{2} - 1) x^{2} + 2 x + 1 + \sqrt{2} = (\sqrt{2} - 1) {(x - x_{0})}^{2} = (\sqrt{2} - 1) {(x + \frac{1}{\sqrt{2} - 1})}^{2}

= (\sqrt{2} - 1) {(x + \sqrt{2} + 1)}^{2} \Rightarrow

F (x) = \frac{a}{x + \sqrt{2} + 1} + \frac{b}{{(x + \sqrt{2} + 1)}^{2}} + \frac{c x + d}{x^{2} + 1} + \frac{e x + f}{{(x^{2} + 1)}^{2}}

b = {l i m}_{x \to - (1 + \sqrt{2})} {(x + \sqrt{2} + 1)}^{2} F (x) \dots . a f t e r f i n d i n g t h e c o e f f i c i e n t s

w e g e t \int F (x) d x = a l n ∣ x + \sqrt{2} + 1 ∣ - \frac{b}{(x + \sqrt{2} + 1)} + \int \frac{c x + d}{x^{2} + 1} d x + \int \frac{e x + f}{{(x^{2} + 1)}^{2}} d x

\dots b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 28/Jan/19

2) l e t f (t) = \int_{0}^{2} \frac{l n (1 + t x)}{1 + x^{2}} d x w i t h t > 0 \Rightarrow

f^{^{'}} (t) = \int_{0}^{2} \frac{x}{(1 + t x) (1 + x^{2})} d x =_{t x = u} \int_{0}^{2 t} \frac{u}{t (1 + u) (1 + \frac{u^{2}}{t^{2}})} \frac{d u}{t}

= \int_{0}^{2 t} \frac{u d u}{(u + 1) (u^{2} + t^{2})} l e t d e c o m p o s e F (u) = \frac{u}{(u + 1) (u^{2} + t^{2})} \Rightarrow

F (u) = \frac{a}{u + 1} + \frac{b u + c}{u^{2} + t^{2}}

a = {l i m}_{u \to - 1} (u + 1) F (u) = - \frac{1}{t^{2} + 1}

{l i m}_{u \to + \infty} u F (u) = 0 = a + b \Rightarrow b = \frac{1}{t^{2} + 1}

F (0) = 0 = a + \frac{c}{t^{2}} \Rightarrow c = - {a t}^{2} = \frac{t^{2}}{t^{2} + 1} \Rightarrow

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

= \frac{1}{t^{2} + 1} {- \frac{1}{u + 1} + \frac{u + t^{2}}{u^{2} + t^{2}}} \Rightarrow \int_{0}^{2 t} F (u) d u

= \frac{1}{t^{2} + 1} {- \int_{0}^{2 t} \frac{d u}{u + 1} + \int_{0}^{2 t} \frac{u + t^{2}}{u^{2} + t^{2}} d u}

= \frac{1}{t^{2} + 1} {- {[l n ∣ u + 1 ∣]}_{0}^{2 t} + \frac{1}{2} {[l n (u^{2} + t^{2})]}_{0}^{2 t} + 2 t^{2} \int_{0}^{2 t} \frac{d u}{u^{2} + t^{2}}}

= \frac{1}{t^{2} + 1} {- l n (1 + 2 t) + \frac{1}{2} (l n (5 t^{2}) - l n (t^{2})) + 2 t^{2} \int_{0}^{2 t} \frac{d u}{u^{2} + t^{2}}}

= - \frac{l n (1 + 2 t)}{t^{2} + 1} + \frac{l n (5)}{2 (t^{2} + 1)} + \frac{2 t^{2}}{t^{2} + 1} \int_{0}^{2 t} \frac{d u}{u^{2} + t^{2}} b u t

\int_{0}^{2 t} \frac{d u}{u^{2} + t^{2}} =_{u = t α} \int_{0}^{2} \frac{t d α}{t^{2} (1 + α^{2})} = \frac{1}{t} \int_{0}^{2} \frac{d α}{1 + α^{2}} = \frac{a r c t a n (2)}{t} \Rightarrow

f^{^{'}} (t) = - \frac{l n (1 + 2 t)}{t^{2} + 1} + \frac{l n (5)}{2 (t^{2} + 1)} + \frac{2 a r c t a n (2) t}{t^{2} + 1} \Rightarrow

f (t) = - \int_{0}^{t} \frac{l n (1 + 2 x)}{1 + x^{2}} d x + \frac{l n (5)}{2} \int_{0}^{t} \frac{d x}{1 + x^{2}} + a r c t a n (2) \int_{0}^{t} \frac{2 x}{1 + x^{2}} d x + c

f (t) = - \int_{0}^{t} \frac{l n (1 + 2 x)}{1 + x^{2}} d x + \frac{l n (5)}{2} a r c t a n (t) + a r c t a n (2) l n (1 + t^{2}) + c

f (0) = 0 \Rightarrow c = 0 \Rightarrow

f (t) = - \int_{0}^{t} \frac{l n (1 + 2 x)}{1 + x^{2}} d x + \frac{l n (5)}{2} a r c t a n (t) + a r c t a n (2) l n (1 + t^{2}) = \int_{0}^{2} \frac{l n (1 + t x)}{1 + x^{2}} d x

w e h a v e f (2) = - \int_{0}^{2} \frac{l n (1 + 2 x)}{1 + x^{2}} d x + \frac{l n (5)}{2} a r c t a n (2) + l n (5) a r c t a n (2)

= \int_{0}^{2} \frac{l n (1 + 2 x)}{1 + x^{2}} d x \Rightarrow 2 \int_{0}^{2} \frac{l n (1 + 2 x)}{1 + x^{2}} d x = \frac{3}{2} l n (5) a r c t a n (2) \Rightarrow

\int_{0}^{2} \frac{l n (1 + 2 x)}{1 + x^{2}} d x = \frac{3}{4} l n (5) a r c t a n (2) . i

Commented by rahul 19 last updated on 29/Jan/19

t h a n k y o u p r o f A b d o .

Commented by rahul 19 last updated on 29/Jan/19

E v e r y o n e p l s t r y Q . 3, 4 a l s o .

Commented by maxmathsup by imad last updated on 29/Jan/19

y o u a r e w e l c o m e s i r .

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19

4) \int_{0}^{π} \frac{(1 + x s i n x) \frac{d u}{d x} - u \frac{d}{d x} (1 + x s i n x)}{{(1 + x s i n x)}^{2}} d x

\int_{0}^{π} \frac{d}{d x} (\frac{u}{1 + x s i n x}) d x

\int_{0}^{π} d (\frac{u}{1 + x s i n x})

∣ \frac{u}{1 + x s i n x} ∣_{0}^{π} u = g (x) s a y

= \frac{g (π)}{1 + π s i n π} - \frac{g (0)}{1 + 0 \times s i n 0}

= g (π) - g (0)

w a i t i h a v e t o f i n d g (x) = u \dots