∫_0 ^(π/2) sin x (√(sin 2x)) dx=? ∫_(−(π/4)) ^(π/4) cos x (√(cos 2x)) dx=?


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Question Number 52900 by MJS last updated on 15/Jan/19

$\underset{0}{\int^{\frac{π}{2}}} \sin x \sqrt{\sin 2 x} d x = ?$ $\underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \cos x \sqrt{\cos 2 x} d x = ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19
	$I=∫_0 ^(π/2) sinx(√(sin2x)) dx I=∫_0 ^(π/2) cosx(√(sin2x)) dx [∫_0 ^a f(x)dx=∫_0 ^a f(a−x)dx] 2I=∫_0 ^(π/2) (cosx+sinx)(√(sin2x)) dx =∫_0 ^(π/2) d(sinx−cosx)(√(1−(1−sin2x))) dx ∫_0 ^(π/2) d(sinx−cosx)(√(1−(sinx−cosx)^2 )) dx ∣(((sinx−cosx)(√(1−(sinx−cosx)^2 )))/2)+(1/2)sin^(−1) (((sinx−cosx)/1))∣_0 ^(π/2) =[{(((1−0)(√(1−(1−0)^2 )))/2)+(1/2)sin^(−1) (((1−0)/1))}−{((0−1(√(1−1)))/2)+(1/2)sin^(−1) (((0−1)/1))}] =(1/2)×(π/2)−{(1/2)×(((−π)/2))} =(π/4)+(π/4)=(π/2) so I=(1/2)×(π/2)=(π/4) sir pls check...$
	$I = \int_{0}^{\frac{π}{2}} s i n x \sqrt{s i n 2 x} d x$ $I = \int_{0}^{\frac{π}{2}} c o s x \sqrt{s i n 2 x} d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$ $2 I = \int_{0}^{\frac{π}{2}} (c o s x + s i n x) \sqrt{s i n 2 x} d x$ $= \int_{0}^{\frac{π}{2}} d (s i n x - c o s x) \sqrt{1 - (1 - s i n 2 x)} d x$ $\int_{0}^{\frac{π}{2}} d (s i n x - c o s x) \sqrt{1 - {(s i n x - c o s x)}^{2}} d x$ $∣ \frac{(s i n x - c o s x) \sqrt{1 - {(s i n x - c o s x)}^{2}}}{2} + \frac{1}{2} {s i n}^{- 1} (\frac{s i n x - c o s x}{1}) ∣_{0}^{\frac{π}{2}}$ $= [{\frac{(1 - 0) \sqrt{1 - {(1 - 0)}^{2}}}{2} + \frac{1}{2} {s i n}^{- 1} (\frac{1 - 0}{1})} - {\frac{0 - 1 \sqrt{1 - 1}}{2} + \frac{1}{2} {s i n}^{- 1} (\frac{0 - 1}{1})}]$ $= \frac{1}{2} \times \frac{π}{2} - {\frac{1}{2} \times (\frac{- π}{2})}$ $= \frac{π}{4} + \frac{π}{4} = \frac{π}{2}$ $s o I = \frac{1}{2} \times \frac{π}{2} = \frac{π}{4} s i r p l s c h e c k . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

	$\int_{- \frac{π}{4}}^{\frac{π}{4}} c o s x \sqrt{c o s 2 x} d x$ $\int_{- a}^{a} f (x) d x = 2 \times \int_{0}^{a} f (x) d x [h e r e f (x) = e v e n f u n c t i o n]$ $s o \int_{\frac{- π}{4}}^{\frac{π}{4}} c o s x \sqrt{c o s 2 x} d x = 2 \times \int_{0}^{\frac{π}{4}} c o s x \sqrt{c o s 2 x} d x$ $n o w u s i n g h e l p o f g r a p h . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

	Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

	$s o \int_{\frac{- π}{4}}^{\frac{π}{4}} c o s x \sqrt{c o s 2 x} d x = a r e a u n d e r t h e c u r v e$ $\approx \frac{2}{3} \times \frac{π}{2} \times 1 = \frac{π}{3}$ $r o u g h l y i t i s a p a r a b o l a s i r . . . s o a r e a$ $\frac{2}{3} a b$
	Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

	Answered by mr W last updated on 16/Jan/19

	$\underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \cos x \sqrt{\cos 2 x} d x$ $= 2 \underset{0}{\int^{\frac{π}{4}}} \cos x \sqrt{\cos 2 x} d x$ $= 2 \underset{0}{\int^{\frac{π}{4}}} \cos x \sqrt{1 - 2 \sin^{2} x} d x$ $= \sqrt{2} \underset{0}{\int^{\frac{π}{4}}} \sqrt{1 - {(\sqrt{2} \sin x)}^{2}} d (\sqrt{2} \sin x) (\int \sqrt{1 - t^{2}} d t)$ $= \frac{\sqrt{2}}{2} {[\sin^{- 1} (\sqrt{2} \sin x) + (\sqrt{2} \sin x) \sqrt{1 - {(\sqrt{2} \sin x)}^{2}}]}_{0}^{\frac{π}{4}}$ $= \frac{\sqrt{2}}{2} (\frac{π}{2})$ $= \frac{π \sqrt{2}}{4}$
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