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Question Number 86399 by Power last updated on 28/Mar/20

Commented by Prithwish Sen 1 last updated on 28/Mar/20

$put x = {2 asin}^{2} t$ $and simplify .$
Commented by mathmax by abdo last updated on 28/Mar/20
$I =∫ ((x(√x))/(√(2a−x))) dx we do the changement (√(2a−x))=t ⇒2a−x=t^2 ⇒ x=2a−t^2 ⇒ I =∫ (((2a−t^2 )(√(2a−t^2 )))/t)(−2t)dt =−2 ∫ (2a−t^2 )(√(2a−t^2 ))dt =_(t=(√(2a))sinu) −2∫ (2a−2asin^2 u)(√(2a))cosu (√(2a))cosu du =−4a(2a) ∫ (1−sin^2 u)cos^2 u du =−8a^2 ∫ cos^4 u du =−8a^2 ∫ (((1+cos(2u))/2))^2 du =−4a^2 ∫ (1+2cos(2u) +cos^2 (2u))du =−4a^2 { u +sin(2u)}−4a^2 ∫ ((1+cos(4u))/2)du =−4a^2 u −4a^2 sin(2u) −2a^2 (u +(1/4)sin(4u)) +C =−6a^2 u −4a^2 sin(2u)−(a^2 /2)sin(4u)+C =−6a^2 u −8a^2 sinu cosu −a^2 sin(2u)cos(2u) +C =−6a^2 u −8a^2 sinucosu −2a^2 sinu cosu(1−2sin^2 u) +C =−2a^2 { 3 arcsin((t/(√(2a))))+4((t/(√(2a))))(√(1−((t/(√(2a))))^2 )) +((t/(√(2a))))(√(1−((t/(√(2a))))^2 ))(1−2((t/(√(2a))))^2 ) }+C I=−2a^2 {3arcsin(((√(2a−x))/(√(2a))))+((4((√(2a−x))))/(√(2a)))(√(1−(((√(2a−x))/(√(2a))))^2 )) +((√(2a−x))/(√(2a)))(√(1−(((√(2a−x))/(√(2a))))^2 ))(1−2(((√(2a−x))/(√(2a))))^2 } +C$
$I = \int \frac{x \sqrt{x}}{\sqrt{2 a - x}} d x w e d o t h e c h a n g e m e n t \sqrt{2 a - x} = t \Rightarrow 2 a - x = t^{2} \Rightarrow$ $x = 2 a - t^{2} \Rightarrow I = \int \frac{(2 a - t^{2}) \sqrt{2 a - t^{2}}}{t} (- 2 t) d t$ $= - 2 \int (2 a - t^{2}) \sqrt{2 a - t^{2}} d t =_{t = \sqrt{2 a} s i n u} - 2 \int (2 a - 2 {a s i n}^{2} u) \sqrt{2 a} c o s u \sqrt{2 a} c o s u d u$ $= - 4 a (2 a) \int (1 - {s i n}^{2} u) {c o s}^{2} u d u$ $= - 8 a^{2} \int {c o s}^{4} u d u = - 8 a^{2} \int {(\frac{1 + c o s (2 u)}{2})}^{2} d u$ $= - 4 a^{2} \int (1 + 2 c o s (2 u) + {c o s}^{2} (2 u)) d u$ $= - 4 a^{2} {u + s i n (2 u)} - 4 a^{2} \int \frac{1 + c o s (4 u)}{2} d u$ $= - 4 a^{2} u - 4 a^{2} s i n (2 u) - 2 a^{2} (u + \frac{1}{4} s i n (4 u)) + C$ $= - 6 a^{2} u - 4 a^{2} s i n (2 u) - \frac{a^{2}}{2} s i n (4 u) + C$ $= - 6 a^{2} u - 8 a^{2} s i n u c o s u - a^{2} s i n (2 u) c o s (2 u) + C$ $= - 6 a^{2} u - 8 a^{2} s i n u c o s u - 2 a^{2} s i n u c o s u (1 - 2 {s i n}^{2} u) + C$ $= - 2 a^{2} {3 a r c s i n (\frac{t}{\sqrt{2 a}}) + 4 (\frac{t}{\sqrt{2 a}}) \sqrt{1 - {(\frac{t}{\sqrt{2 a}})}^{2}}$ $+ (\frac{t}{\sqrt{2 a}}) \sqrt{1 - {(\frac{t}{\sqrt{2 a}})}^{2}} (1 - 2 {(\frac{t}{\sqrt{2 a}})}^{2})} + C$ $I = - 2 a^{2} {3 a r c s i n (\frac{\sqrt{2 a - x}}{\sqrt{2 a}}) + \frac{4 (\sqrt{2 a - x})}{\sqrt{2 a}} \sqrt{1 - {(\frac{\sqrt{2 a - x}}{\sqrt{2 a}})}^{2}}$ $+ \frac{\sqrt{2 a - x}}{\sqrt{2 a}} \sqrt{1 - {(\frac{\sqrt{2 a - x}}{\sqrt{2 a}})}^{2}} (1 - 2 {(\frac{\sqrt{2 a - x}}{\sqrt{2 a}})}^{2}} + C$
Commented by jagoll last updated on 28/Mar/20

$standard solving$ $let a - x = a \cos 2 t$ $\frac{x}{2 a - x} = \frac{1 + \cos 2 t}{1 - \cos 2 t} = \cot^{2} t$ $\int {2 a}^{2} \times a (1 - \cos 2 t) \sin 2 t \cot t dt$ $= {8 a}^{2} \int \sin^{2} t \cos^{2} t dt$ $= {2 a}^{2} \int \sin^{2} 2 t dt$ $= {2 a}^{2} \int (\frac{1}{2} - \frac{1}{2} \cos 4 t) dt$ $= a^{2} t - \frac{a^{2}}{4} \sin 4 t + c$ $super easy to solve$
	Answered by john santu last updated on 28/Mar/20

	Commented by Power last updated on 28/Mar/20

	$thank you sir$
	Commented by peter frank last updated on 29/Mar/20

	$t h a n k y o u b o t h$
	Answered by MJS last updated on 28/Mar/20

	$\int x \sqrt{\frac{x}{2 a - x}} d x =$ $[t = \sqrt{\frac{x}{2 a - x}} \to d x = \frac{4 a t}{{(t^{2} + 1)}^{2}} d t]$ $= 8 a^{2} \int \frac{t^{4}}{{(t^{2} + 1)}^{3}} d t =$ $= - \frac{a^{2} t (5 t^{2} + 3)}{{(t^{2} + 1)}^{3}} + 3 a^{2} \arctan t =$ $= - \frac{(x + 3 a) \sqrt{2 a x - x^{2}}}{2} + 3 a^{2} \arctan \sqrt{\frac{x}{2 a - x}} + C$
	Commented by jagoll last updated on 28/Mar/20

	$super easy sir ?$
	Commented by MJS last updated on 28/Mar/20

	$well well well . . .$
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