0-pi-2-sin-x-sin-2x-dx-pi-4-pi-4-cos-x-cos-2x-dx-

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 = \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 \dots

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 \dots

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 \dots 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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