∫_( 0) ^( a/2) x^2 (a^2 − x^2 )^(3/2) dx


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Question Number 17603 by tawa tawa last updated on 08/Jul/17

$\int_{0}^{a / 2} x^{2} {(a^{2} - x^{2})}^{\frac{3}{2}} dx$
	Answered by alex041103 last updated on 08/Jul/17

	$I = \int_{0}^{a / 2} x^{2} {(a^{2} - x^{2})}^{\frac{3}{2}} dx$ $L e t x = a s i n θ (i . e . θ = {s i n}^{- 1} (\frac{x}{a}))$ $\Rightarrow d x = a c o s θ$ $W e u s e 1 - {s i n}^{2} θ = {c o s}^{2} θ a n d$ $θ = {s i n}^{- 1} (\frac{x}{a}) t o c h a n g e t h e l i m i t s o f$ $i n t e g r a t i o n :$ $I = a^{6} \underset{0}{\int^{π / 6}} {s i n}^{2} θ {c o s}^{4} θ d θ$ $T h e n w e s u b s t i t u t e {s i n}^{2} θ = 1 - {c o s}^{2} θ$ $I = a^{6} [\underset{0}{\int^{π / 6}} {c o s}^{4} θ d θ - \underset{0}{\int^{π / 6}} {c o s}^{6} θ d θ]$ $L e t I_{n} = \underset{0}{\int^{π / 6}} {c o s}^{n} θ d θ$ $\Rightarrow I = a^{6} (I_{4} - I_{6})$ $I_{n} = \underset{0}{\int^{π / 6}} {c o s}^{n - 1} θ c o s θ d θ$ $T h e n w e u s e i n t e g r a t i o n b y p a r t s :$ $u = {c o s}^{n - 1} θ d v = c o s θ d θ$ $d u = (n - 1) (- s i n θ) {c o s}^{n - 2} θ v = s i n θ$ $\underset{a}{\int^{b}} u d v = {[u v]}_{a}^{b} - \underset{a}{\int^{b}} v d u$ $I_{n} = {[{c o s}^{n - 1} θ s i n θ]}_{0}^{π / 6} + (n - 1) \underset{0}{\int^{π / 6}} {s i n}^{2} θ {c o s}^{n - 2} θ d θ$ $W e k n o w t h a t {s i n}^{2} θ = 1 - {c o s}^{2} θ$ $\Rightarrow I_{n} = \frac{3^{\frac{1}{2} (n - 1)}}{2^{n}} + (n - 1) (\underset{0}{\int^{π / 6}} {c o s}^{n - 2} θ d θ - \underset{0}{\int^{π / 6}} {c o s}^{n} θ d θ)$ $B u t$ $I_{n} = \underset{0}{\int^{π / 6}} {c o s}^{n} θ d θ$ $I_{n - 2} = \underset{0}{\int^{π / 6}} {c o s}^{n - 2} θ d θ$ $\Rightarrow I_{n} = \frac{3^{\frac{n - 1}{2}}}{2^{n}} + (n - 1) I_{n - 2} - (n - 1) I_{n}$ $i . e . I_{n} = \frac{3^{\frac{n - 1}{2}}}{n 2^{n}} + \frac{n - 1}{n} I_{n - 2}$ $W e u s e t h e f o r m u l a f o r I = a^{6} (I_{6 - 2} - I_{6})$ $a n d w e g e t$ $I = a^{6} (I_{6 - 2} - (\frac{3 \sqrt{3}}{2^{7}} + \frac{5}{6} I_{6 - 2}))$ $= a^{6} (\frac{1}{6} I_{4} - \frac{3 \sqrt{3}}{2^{7}})$ $N o w w d u s e t h e f o r m u l a f o r I_{n} t w i c e$ $t o e v a l u a t e I_{4} (i . e . e x p r e s s i n g I_{4} i n t e r m s o f I_{0}) a n d w e g e t$ $I_{4} = \frac{3 \sqrt{3}}{2^{6}} + \frac{3 \sqrt{3}}{2^{5}} + \frac{π}{2^{4}}$ $N o w w e f i n a l y g e t f o r I :$ $I = \underset{0}{\int^{a / 2}} x^{2} {(a^{2} - x^{2})}^{3 / 2} d x = \frac{a^{6} π}{96}$
	Commented by tawa tawa last updated on 08/Jul/17

	$God bless you sir . i really appreciate .$
	Answered by Arnab Maiti last updated on 13/Jul/17
	$let I =∫_0 ^( (a/2)) x^2 (a^2 −x^2 )^(3/2) dx put x=a sinθ ∴ dx= a cosθ dθ ∴ I=∫_0 ^( (π/6)) a^3 sin^2 θ{a^2 (1−sin^2 θ)}^(3/2) cosθ dθ =a^6 ∫_0 ^(π/6) sin^2 θ cos^4 θ dθ =a^6 (1/8)∫_0 ^( (π/6)) (2sinθ cosθ)^2 2cos^2 θ dθ =a^6 ×(1/8)∫_0 ^(π/6) sin^2 2θ (1+cos2θ)dθ =a^6 ×(1/8)[∫_0 ^(π/6) sin^2 2θ dθ+∫_0 ^(π/6) sin^2 2θ cos2θ dθ] =a^6 /8×[(1/2)∫_0 ^(π/6) (1−cos4θ)dθ+(1/2)∫_0 ^((√3)/2) (sin 2θ)^2 d(sin 2θ)] =(a^6 /(16))[ θ−((sin 4θ)/4)]_0 ^(π/6) +(a^6 /(16))[(((p)^3 )/3)]_0 ^((√3)/2) let sin2θ=p =(a^6 /(16))[{(π/6)−((√3)/8)}+((3(√3))/(3×8))] =(a^6 /(16))×(π/6)=((πa^6 )/(96))$
	$let I = \int_{0}^{\frac{a}{2}} x^{2} {(a^{2} - x^{2})}^{\frac{3}{2}} dx$ $put x = a \sin θ ∴ dx = a \cos θ d θ$ $∴ I = \int_{0}^{\frac{π}{6}} a^{3} \sin^{2} θ {a^{2} (1 - \sin^{2} θ)}^{\frac{3}{2}} \cos θ d θ$ $= a^{6} \int_{0}^{\frac{π}{6}} \sin^{2} θ \cos^{4} θ d θ$ $= a^{6} \frac{1}{8} \int_{0}^{\frac{π}{6}} {(2 \sin θ \cos θ)}^{2} {2 \cos}^{2} θ d θ$ $= a^{6} \times \frac{1}{8} \int_{0}^{\frac{π}{6}} \sin^{2} 2 θ (1 + \cos 2 θ) d θ$ $= a^{6} \times \frac{1}{8} [\int_{0}^{\frac{π}{6}} \sin^{2} 2 θ d θ + \int_{0}^{\frac{π}{6}} \sin^{2} 2 θ \cos 2 θ d θ]$ $= a^{6} / 8 \times [\frac{1}{2} \int_{0}^{\frac{π}{6}} (1 - \cos 4 θ) d θ + \frac{1}{2} \int_{0}^{\frac{\sqrt{3}}{2}} {(\sin 2 θ)}^{2} d (\sin 2 θ)]$ $= \frac{a^{6}}{16} {[θ - \frac{\sin 4 θ}{4}]}_{0}^{\frac{π}{6}} + \frac{a^{6}}{16} {[\frac{{(p)}^{3}}{3}]}_{0}^{\frac{\sqrt{3}}{2}} let \sin 2 θ = p$ $= \frac{a^{6}}{16} [{\frac{π}{6} - \frac{\sqrt{3}}{8}} + \frac{3 \sqrt{3}}{3 \times 8}]$ $= \frac{a^{6}}{16} \times \frac{π}{6} = \frac{π a^{6}}{96}$
	Commented by tawa tawa last updated on 11/Jul/17

	$God bless you sir .$
	Commented by alex041103 last updated on 11/Jul/17
	$You have a mistake x=a sin θ →dx=cos θ dθ ∴ I=∫_0 ^( (π/6)) a^3 sin^2 θ{a^2 (1−sin^2 θ)}^(3/2) cos θ dθ I=a^6 ∫_( 0) ^(π/6) sin^2 θ cos^4 θ dθ ≠ a^6 ∫_( 0) ^(π/6) sin^2 θ cos^3 θ dθ$
	$You have a mistake$ $x = a \sin θ \to d x = \cos θ d θ$ $∴ I = \int_{0}^{\frac{π}{6}} a^{3} \sin^{2} θ {a^{2} (1 - \sin^{2} θ)}^{\frac{3}{2}} \cos θ d θ$ $I = a^{6} \underset{0}{\int^{π / 6}} \sin^{2} θ \cos^{4} θ d θ \neq a^{6} \underset{0}{\int^{π / 6}} \sin^{2} θ \cos^{3} θ d θ$
	Commented by Arnab Maiti last updated on 12/Jul/17

	$Now it is corrected . Still it is different answer .$
	Commented by alex041103 last updated on 13/Jul/17

	$Y o u n o w h a v e a n o t h e r m i s t a k e$ $A t l a s t l i n e s w h e r e y o u f i n i s h t h e$ $i n t e g r a t i o n y o u s e t {s i n}^{2} 2 θ = \frac{1 - c o s 2 θ}{2}$ $w h i c h i s n o t t r u e$ $\frac{1 - c o s 2 θ}{2} = {s i n}^{2} θ \neq {s i n}^{2} 2 θ$
	Commented by Arnab Maiti last updated on 13/Jul/17

	$The answer is indentical now .$ $Thank you!!$
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