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Question Number 37938 by Fawomath last updated on 19/Jun/18

Commented by abdo.msup.com last updated on 19/Jun/18
$1) let I_p = ∫ x^2 (n−x)^p dx by parts I_p = −(1/(p+1))x^2 (n−x)^(p+1) +(1/(p+1))∫ 2x (n−x)^(p+1) dx =−(x^2 /(p+1))(n−x)^(p+1) + +(2/(p+1)){ −(1/(p+2))x(n−x)^(p+2) +(1/(p+2))∫ (n−x)^(p+2) dx} =−(x^2 /(p+1))(n−x)^(p+1) −((2x)/((p+1)(p+2)))(n−x)^(p+2) −(2/((p+1)(p+2)(p+3)))(n−x)^(p+3) +c$
$1) l e t I_{p} = \int x^{2} {(n - x)}^{p} d x b y p a r t s$ $I_{p} = - \frac{1}{p + 1} x^{2} {(n - x)}^{p + 1} + \frac{1}{p + 1} \int 2 x {(n - x)}^{p + 1} d x$ $= - \frac{x^{2}}{p + 1} {(n - x)}^{p + 1} +$ $+ \frac{2}{p + 1} {- \frac{1}{p + 2} x {(n - x)}^{p + 2} + \frac{1}{p + 2} \int {(n - x)}^{p + 2} d x}$ $= - \frac{x^{2}}{p + 1} {(n - x)}^{p + 1} - \frac{2 x}{(p + 1) (p + 2)} {(n - x)}^{p + 2}$ $- \frac{2}{(p + 1) (p + 2) (p + 3)} {(n - x)}^{p + 3} + c$
Commented by abdo.msup.com last updated on 19/Jun/18

$2) \int \frac{x^{2}}{x + 1} d x = \int \frac{x^{2} - 1 + 1}{x + 1} d x$ $= \int (x - 1) d x + \int \frac{d x}{x + 1}$ $= \frac{x^{2}}{2} - x + l n ∣ x + 1 ∣ + c .$
Commented by abdo.msup.com last updated on 19/Jun/18

$3) l e t I = \int \frac{a r c s i n x}{\sqrt{1 - x^{2}}} d x b y p a r t s$ $I = a r c s i n x . a r c s i n x - \int \frac{a r c s i n x}{\sqrt{1 - x^{2}}} \Rightarrow$ $2 I = {(a r c s i n x)}^{2} \Rightarrow I = \frac{1}{2} {(a r c s i n x)}^{2} + c .$
Commented by abdo.msup.com last updated on 19/Jun/18

$5) \int \frac{s i n x}{{(1 + c o s x)}^{2}} d x = \int \frac{- d (c o s x)}{{(1 + c o s x)}^{2}}$ $= \frac{- 1}{1 + c o s x} + c .$
Commented by abdo.msup.com last updated on 19/Jun/18

$4) w e c a n p r o v e t h i s e q u a l i t y b y$ $r e c u r r e n c e a l s o w e c a n s e a r c h$ $a p o l y n o m a t f o r m$ $p (x) = {a x}^{6} + {b x}^{5} + c x^{4} + {d x}^{3} + {e x}^{2} + f x$ $w i c h v e r i f y p (x + 1) - p (x) = x^{5} a n d p (0) = 0 s o$ $\sum_{k = 1}^{n} k^{5} = \sum_{k = 1}^{n} p (k + 1) - p (k)$ $= p (n + 1) - p (1) . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18
	$1)let t=n−x dt=−dx ∫(n−t)^2 ×t^p ×−dt −∫(n^2 t^p −2nt^(p+1) +t^(p+2) )dt −{n^2 (t^(p+1) /(p+1))−2n(t^(p+2) /(p+2))+(t^(p+3) /(p+3))}+c −{n^2 (((n−x)^(p+1) )/(p+1))2n(((n−x)^(p+2) )/(p+2))+(((n−x)^(p+3) )/(p+3))}+cc$
	$1) l e t t = n - x d t = - d x$ $\int {(n - t)}^{2} \times t^{p} \times - d t$ $- \int (n^{2} t^{p} - 2 {n t}^{p + 1} + t^{p + 2}) d t$ $- {n^{2} \frac{t^{p + 1}}{p + 1} - 2 n \frac{t^{p + 2}}{p + 2} + \frac{t^{p + 3}}{p + 3}} + c$ $- {n^{2} \frac{{(n - x)}^{p + 1}}{p + 1} 2 n \frac{{(n - x)}^{p + 2}}{p + 2} + \frac{{(n - x)}^{p + 3}}{p + 3}} + c c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

	$2) \int \frac{x^{2} - 1 + 1}{x + 1} d x$ $\int x - 1 d x + \int \frac{d x}{x + 1}$ $\frac{x^{2}}{2} - x + l n (x + 1) + c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

	$3) \int \frac{{s i n}^{- 1} x}{\sqrt{1 - x^{2}}} d x$ $t = {s i n}^{- 1} x d t = \frac{d x}{\sqrt{1 - x^{2}}}$ $\int t d t$ $\frac{t^{2}}{2} + c$ $= \frac{1}{2} {({s i n}^{- 1} x)}^{2} + c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

	$\int {t a n}^{2} {x s e c}^{2} x d x$ $t = t a n x d t = {s e c}^{2} x d x$ $\int t^{2} d t$ $\frac{t^{3}}{3} + c$ $\frac{{(t a n x)}^{3}}{3} + c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

	$5) \int \frac{s i n x}{{(1 + c o s x)}^{2}} d x$ $= - 1 \int \frac{d (1 + c o s x}{{1 + c o s x)}^{2}} \overset{}{}$ $= - 1 \times \frac{1}{1 + c o s x} \times - 1 = \frac{1}{1 + c o s x}$
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