Evaluate : 1) ∫_(−1) ^( 1) cot^(−1) ((1/(√(1−x^2 )))).(cot^(−1) (x/(√(1−(x^2 )^(∣x∣) ))))dx 2) ∫_0 ^( (π/2)) ((sin^2 (10)θ)/(sin^2 θ)) dθ 3) ∫_0 ^( (π/4)) ((ln(cotx))/(((sinx)^(2009) +(cosx)^(2009) )^2 )).(sin2x)^(2008) dx 4) ∫


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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

$N o p r o f . U a r e w r o n g . . .$ $I t {a i n}^{'} t a n o d d f u n c t i o n s i n c e$ ${c o t}^{- 1} (- x) = π - \cot^{- 1} (x) f o r a l l x ϵ R .$ $A n s : \frac{π^{2} (\sqrt{2} - 1)}{2} .$ $I n d e e d a v e r y b e a u t i f u l p r o b l e m .$
Commented by Meritguide1234 last updated on 30/Jan/19

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

$l e t a r c c o t a n (t) = y \Leftrightarrow t = c o t a n (y) = \frac{1}{t a n y} \Rightarrow t a n y = \frac{1}{t} \Rightarrow y = a r c t a n (\frac{1}{t})$ $= \bar{+} \frac{π}{2} - a r c t a n (t) \Rightarrow a r c o t a n (t) = \bar{+} \frac{π}{2} - a r c t a n (t) \Rightarrow$ $a r c o t a n (- t) = \bar{+} \frac{π}{2} + a r c t a n (t) y o u a r e r i g h t s i r t h e f u n c t i o n i s n o t o d d . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19

	$2) \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} n θ}{{s i n}^{2} θ} d θ [n = 10]$ $I_{n} = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{1 - c o s 2 n θ}{{s i n}^{2} θ} d θ$ $I_{n} - I_{n - 2} = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{c o s (2 n - 2) θ - c o s 2 n θ}{{s i n}^{2} θ} d θ$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{2 s i n (2 n - 1) θ s i n θ}{{s i n}^{2} θ} d θ$ $= \int_{0}^{\frac{π}{2}} \frac{s i n (2 n - 1) θ}{s i n θ} d θ$ $J_{n} = I_{n} - I_{n - 1}$ $J_{n} - J_{n - 1} = \int_{0}^{\frac{π}{2}} \frac{s i n (2 n - 1) θ - s i n (2 n - 3) θ}{s i n θ} d θ$ $= \int_{0}^{\frac{π}{2}} \frac{2 c o s (2 n - 2) θ s i n θ}{s i n θ} d θ$ $= 2 ∣ \frac{s i n (2 n - 2) θ}{2 n - 2} ∣_{0}^{\frac{π}{2}} = 0$ $s o J_{n} - J_{n - 1} = 0$ $J_{n} = J_{n - 1}$ $I_{n} - I_{n - 1} = I_{n - 1} - I_{n - 2}$ $I_{n} = 2 I_{n - 1} - I_{n - 2}$ $\int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} (n θ)}{{s i n}^{2} θ} = I_{n} w e h a v e t o f i n d I_{10}$ $I_{n} = 2 I_{n - 1} - I_{n - 2} [I_{0} = 0]$ $I_{2} = 2 I_{1} - I_{0} = 2 I_{1} - 0 = 2 \times \frac{π}{2} = π [I_{1} = \frac{π}{2}]$ $I_{3} = 2 I_{2} - I_{1} = \frac{3 π}{2}$ $I_{4} = 2 π$ $. . . . .$ $t h u s o n c a l c u l a t i o n (a t t a c h e d p h o t o) w e g e t$ $I_{10} = 5 π$ $A g e n e r a l f o r m u l a t h u s o b t a i n e d$ $\int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} (n θ)}{{s i n}^{2} θ} d θ = n (\frac{π}{2}) = I_{n}$
	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

	$3) \int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{[{(s i n x)}^{2009} + {(c o s x)}^{2009}]}^{2}} {(s i n 2 x)}^{2008} d x$ $\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{[{(s i n x)}^{a} + {(c o s x)}^{a}]}^{2}} \times {(2 s i n x c o s x)}^{a - 1} d x$ $\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{(c o s x)}^{2 a} {[1 + {(t a n x)}^{a}]}^{2}} \times {(2 {t a n x c o s}^{2} x)}^{a - 1} d x$ $\int_{0}^{\frac{π}{4}} \frac{- l n (t a n x)}{{(c o s x)}^{2 a} {[1 + {(t a n x)}^{a}]}^{2}} \times {(2)}^{a - 1} \times {(t a n x)}^{a - 1} \times {(c o s x)}^{2 a - 2} d x$ $= 2^{a - 1} \int_{0}^{\frac{π}{4}} \frac{- l n (t a n x)}{{[1 + {(t a n x)}^{a}]}^{2}} \times {s e c}^{2} x \times {(t a n x)}^{a - 1} d x$ $l e t k = {(t a n x)}^{a} \frac{d k}{d x} = a {(t a n x)}^{a - 1} \times {s e c}^{2} x$ $\frac{d k}{a} = {(t a n x)}^{a - 1} \times {s e c}^{2} x d x$ $l n k = a (l n t a n x)$ $= 2^{a - 1} \int_{0}^{1} \frac{\frac{- l n k}{a}}{{[1 + k]}^{2}} \times \frac{d k}{a}$ $= - 2^{a - 1} \int_{0}^{1} \frac{l n k}{{[1 + k]}^{2}} d k$ $n o w \int \frac{l n k}{{(1 + k)}^{2}}$ $l n k \int \frac{d k}{{(1 + k)}^{2}} - [\frac{d}{d k} (l n k) \int \frac{d k}{{(1 + k)}^{2}}] d k$ $= \frac{- l n k}{(1 + k)} - \int \frac{1}{k} \times \frac{- 1}{1 + k} d k$ $= \frac{- l n k}{(1 + k)} + \int \frac{k + 1 - k}{k (k + 1)} d k$ $= \frac{- l n k}{(1 + k)} + \int \frac{d k}{k} - \int \frac{d k}{k + 1}$ $= \frac{- l n k}{(1 + k)} + l n k - l n (k + 1)$ $= \frac{- l n k}{(1 + k)} + l n (\frac{k}{k + 1}) + c$ $s o a n s w e r o f (- 2^{a - 1}) \int_{0}^{1} \frac{l n k}{{(1 + k)}^{2}} d k i s$ $= (- 2^{a - 1}) ∣ \frac{- l n k}{1 + k} + l n (\frac{k}{k + 1}) ∣_{0}^{1}$ $= (- 2^{a - 1}) [(\frac{- l n 1}{1 + 1} + l n (\frac{1}{2}) + \frac{l n (0)}{1 + 0} - l n (0)]$ $s o i n t h i n k l n (0) - l n (0) c a n c e l l e d$ $a n d a n s w e r i s$ $= (- 2^{a - 1}) \times (- l n 2) = 2^{a - 1} l n 2 = 2^{2008} l n 2 i s a n s w e r$
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