Evaluate-1-1-1-cot-1-1-1-x-2-cot-1-x-1-x-2-x-dx-2-0-pi-2-sin-2-10-sin-2-d-3-0-pi-4-ln-cotx-sinx-2009-cosx-200

Question Number 54070 by rahul 19 last updated on 28/Jan/19

E v a l u a t e :

1)

\int_{- 1}^{1} \cot^{- 1} (\frac{1}{\sqrt{1 - x^{2}}}) . (co t^{- 1} \frac{x}{\sqrt{1 - {(x^{2})}^{∣ x ∣}}}) d x

2)

\int_{0}^{\frac{π}{2}} \frac{\sin^{2} (10) θ}{\sin^{2} θ} d θ

3)

\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{({(\sin x)}^{2009} + {(\cos x)}^{2009})}^{2}} . {(\sin 2 x)}^{2008} d x

4)

\int_{0}^{2} \frac{4 x + 10}{{(x^{2} + 5 x + 6)}^{2}} d x .

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

4) \int_{0}^{2} \frac{4 x + 10}{{(x^{2} + 5 x + 6)}^{2}} = 2 \int_{0}^{2} \frac{2 x + 5}{{(x^{2} + 5 x + 6)}^{2}} = {[- 2 \frac{1}{x^{2} + 5 x + 6}]}_{0}^{2}

= - 2 (\frac{1}{20} - \frac{1}{6}) = \frac{1}{3} - \frac{1}{10} = \frac{7}{30} .

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

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

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

Commented by rahul 19 last updated on 30/Jan/19

N o p r o f . U a r e w r o n g \dots

I t {a i n}^{'} t a n o d d f u n c t i o n s i n c e

{c o t}^{- 1} (- x) = π - \cot^{- 1} (x) f o r a l l x ϵ R .

A n s : \frac{π^{2} (\sqrt{2} - 1)}{2} .

I n d e e d a v e r y b e a u t i f u l p r o b l e m .

Commented by Meritguide1234 last updated on 30/Jan/19

Commented by maxmathsup by imad last updated on 30/Jan/19

l e t a r c c o t a n (t) = y \Leftrightarrow t = c o t a n (y) = \frac{1}{t a n y} \Rightarrow t a n y = \frac{1}{t} \Rightarrow y = a r c t a n (\frac{1}{t})

= \bar{+} \frac{π}{2} - a r c t a n (t) \Rightarrow a r c o t a n (t) = \bar{+} \frac{π}{2} - a r c t a n (t) \Rightarrow

a r c o t a n (- t) = \bar{+} \frac{π}{2} + a r c t a n (t) y o u a r e r i g h t s i r t h e f u n c t i o n i s n o t o d d \dots

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

2) \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} n θ}{{s i n}^{2} θ} d θ [n = 10]

I_{n} = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{1 - c o s 2 n θ}{{s i n}^{2} θ} d θ

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

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{2 s i n (2 n - 1) θ s i n θ}{{s i n}^{2} θ} d θ

= \int_{0}^{\frac{π}{2}} \frac{s i n (2 n - 1) θ}{s i n θ} d θ

J_{n} = I_{n} - I_{n - 1}

J_{n} - J_{n - 1} = \int_{0}^{\frac{π}{2}} \frac{s i n (2 n - 1) θ - s i n (2 n - 3) θ}{s i n θ} d θ

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

= 2 ∣ \frac{s i n (2 n - 2) θ}{2 n - 2} ∣_{0}^{\frac{π}{2}} = 0

s o J_{n} - J_{n - 1} = 0

J_{n} = J_{n - 1}

I_{n} - I_{n - 1} = I_{n - 1} - I_{n - 2}

I_{n} = 2 I_{n - 1} - I_{n - 2}

\int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} (n θ)}{{s i n}^{2} θ} = I_{n} w e h a v e t o f i n d I_{10}

I_{n} = 2 I_{n - 1} - I_{n - 2} [I_{0} = 0]

I_{2} = 2 I_{1} - I_{0} = 2 I_{1} - 0 = 2 \times \frac{π}{2} = π [I_{1} = \frac{π}{2}]

I_{3} = 2 I_{2} - I_{1} = \frac{3 π}{2}

I_{4} = 2 π

\dots . .

t h u s o n c a l c u l a t i o n (a t t a c h e d p h o t o) w e g e t

I_{10} = 5 π

A g e n e r a l f o r m u l a t h u s o b t a i n e d

\int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} (n θ)}{{s i n}^{2} θ} d θ = n (\frac{π}{2}) = I_{n}

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

T h a n k y o u s i r!

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

3) \int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{[{(s i n x)}^{2009} + {(c o s x)}^{2009}]}^{2}} {(s i n 2 x)}^{2008} d x

\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{[{(s i n x)}^{a} + {(c o s x)}^{a}]}^{2}} \times {(2 s i n x c o s x)}^{a - 1} d x

\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{(c o s x)}^{2 a} {[1 + {(t a n x)}^{a}]}^{2}} \times {(2 {t a n x c o s}^{2} x)}^{a - 1} d x

\int_{0}^{\frac{π}{4}} \frac{- l n (t a n x)}{{(c o s x)}^{2 a} {[1 + {(t a n x)}^{a}]}^{2}} \times {(2)}^{a - 1} \times {(t a n x)}^{a - 1} \times {(c o s x)}^{2 a - 2} d x

= 2^{a - 1} \int_{0}^{\frac{π}{4}} \frac{- l n (t a n x)}{{[1 + {(t a n x)}^{a}]}^{2}} \times {s e c}^{2} x \times {(t a n x)}^{a - 1} d x

l e t k = {(t a n x)}^{a} \frac{d k}{d x} = a {(t a n x)}^{a - 1} \times {s e c}^{2} x

\frac{d k}{a} = {(t a n x)}^{a - 1} \times {s e c}^{2} x d x

l n k = a (l n t a n x)

= 2^{a - 1} \int_{0}^{1} \frac{\frac{- l n k}{a}}{{[1 + k]}^{2}} \times \frac{d k}{a}

= - 2^{a - 1} \int_{0}^{1} \frac{l n k}{{[1 + k]}^{2}} d k

n o w \int \frac{l n k}{{(1 + k)}^{2}}

l n k \int \frac{d k}{{(1 + k)}^{2}} - [\frac{d}{d k} (l n k) \int \frac{d k}{{(1 + k)}^{2}}] d k

= \frac{- l n k}{(1 + k)} - \int \frac{1}{k} \times \frac{- 1}{1 + k} d k

= \frac{- l n k}{(1 + k)} + \int \frac{k + 1 - k}{k (k + 1)} d k

= \frac{- l n k}{(1 + k)} + \int \frac{d k}{k} - \int \frac{d k}{k + 1}

= \frac{- l n k}{(1 + k)} + l n k - l n (k + 1)

= \frac{- l n k}{(1 + k)} + l n (\frac{k}{k + 1}) + c

s o a n s w e r o f (- 2^{a - 1}) \int_{0}^{1} \frac{l n k}{{(1 + k)}^{2}} d k i s

= (- 2^{a - 1}) ∣ \frac{- l n k}{1 + k} + l n (\frac{k}{k + 1}) ∣_{0}^{1}

= (- 2^{a - 1}) [(\frac{- l n 1}{1 + 1} + l n (\frac{1}{2}) + \frac{l n (0)}{1 + 0} - l n (0)]

s o i n t h i n k l n (0) - l n (0) c a n c e l l e d

a n d a n s w e r i s

= (- 2^{a - 1}) \times (- l n 2) = 2^{a - 1} l n 2 = 2^{2008} l n 2 i s a n s w e r