[reposted] find ∫_( 0) ^( (𝛑/2)) sin^8 (x)dx + ∫_( 0) ^( 1) sin^(-1) ((x)^(1/8) ) dx=?


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Question Number 162471 by mr W last updated on 29/Dec/21

$[r e p o s t e d]$ $f i n d \underset{0}{\int^{\frac{π}{2}}} \sin^{8} (x) d x + \underset{0}{\int^{1}} \sin^{- 1} (\sqrt[8]{x}) d x = ?$
	Answered by Ar Brandon last updated on 30/Dec/21
	$I=∫_0 ^(π/2) sin^8 xdx+∫_0 ^1 sin^(−1) ((x)^(1/8) )dx I_1 =∫_0 ^(π/2) sin^8 xdx , I_2 =∫_0 ^1 sin^(−1) ((x)^(1/8) )dx I_2 =∫_0 ^1 sin^(−1) ((x)^(1/8) )dx , x=t^8 ⇒dx=8t^7 dt =8∫_0 ^1 t^7 sin^(−1) (t)dt, { ((u(t)=sin^(−1) (t))),((v′(t)=t^7 )) :}⇒ { ((u′(t)=(1/( (√(1−t^2 )))))),((v(t)=(1/8)t^8 )) :} =[t^8 sin^(−1) (t)]_0 ^1 −∫_0 ^1 (t^8 /( (√(1−t^2 ))))dt, t=sinϑ⇒dt=cosϑdϑ =(π/2)−∫^(π/2) _0 sin^8 ϑdϑ=(π/2)−∫_0 ^(π/2) sin^8 xdx=(π/2)−I_1 I=I_1 +I_2 =I_1 +(π/2)−I_1 =(π/2)$
	$I = \int_{0}^{\frac{π}{2}} \sin^{8} x d x + \int_{0}^{1} \sin^{- 1} (\sqrt[8]{x}) d x$ $I_{1} = \int_{0}^{\frac{π}{2}} \sin^{8} x d x, I_{2} = \int_{0}^{1} \sin^{- 1} (\sqrt[8]{x}) d x$ $I_{2} = \int_{0}^{1} \sin^{- 1} (\sqrt[8]{x}) d x, x = t^{8} \Rightarrow d x = 8 t^{7} d t$ $= 8 \int_{0}^{1} t^{7} \sin^{- 1} (t) d t, {\begin{cases} u (t) = \sin^{- 1} (t) \\ v^{'} (t) = t^{7} \end{cases} \Rightarrow {\begin{cases} u^{'} (t) = \frac{1}{\sqrt{1 - t^{2}}} \\ v (t) = \frac{1}{8} t^{8} \end{cases}$ $= {[t^{8} \sin^{- 1} (t)]}_{0}^{1} - \int_{0}^{1} \frac{t^{8}}{\sqrt{1 - t^{2}}} d t, t = \sin ϑ \Rightarrow d t = \cos ϑ d ϑ$ $= \frac{π}{2} - {\int_{0}}^{\frac{π}{2}} \sin^{8} ϑ d ϑ = \frac{π}{2} - \int_{0}^{\frac{π}{2}} \sin^{8} x d x = \frac{π}{2} - I_{1}$ $I = I_{1} + I_{2} = I_{1} + \frac{π}{2} - I_{1} = \frac{π}{2}$
	Commented by Ar Brandon last updated on 30/Dec/21

	$Thanks Sir . Corrected! {It}^{'} s okay now .$
	Commented by Ar Brandon last updated on 29/Dec/21

	Commented by Ar Brandon last updated on 29/Dec/21

	$No, Sir .$
	Commented by mr W last updated on 30/Dec/21

	$t h a n k s f o r c o n f i r m i n g s i r!$
	Answered by mr W last updated on 29/Dec/21

	$w e c a n s o l v e t h i s i n t e g r a l w i t h o u t$ $c a l c u l a t i n g i t .$
	Commented by mr W last updated on 29/Dec/21

	$w e w a n t t o f i n d I = I_{1} + I_{2} w i t h$ $I_{1} = \int_{0}^{\frac{π}{2}} \sin^{8} x d x$ $I_{2} = \int_{0}^{1} \sin^{- 1} (\sqrt[8]{x}) d x$ $n o w w e a p p l y m e t h o d a b o v e t o f i n d$ $I_{2} = \int_{0}^{1} \sin^{- 1} (\sqrt[8]{x}) d x$ $y = f (x) = \sin^{- 1} (\sqrt[8]{x})$ $\Rightarrow x = \sin^{8} (y) = f^{- 1} (y)$ $x \in [0, 1] \Rightarrow y \in [0, \frac{π}{2}]$ $I_{2} = \int_{0}^{1} \sin^{- 1} (\sqrt[8]{x}) d x$ $= 1 \times \frac{π}{2} - \int_{0}^{\frac{π}{2}} \sin^{8} (y) d y$ $= \frac{π}{2} - \int_{0}^{\frac{π}{2}} \sin^{8} (x) d x$ $= \frac{π}{2} - I_{1}$ $I = I_{1} + I_{2} = I_{1} + \frac{π}{2} - I_{1} = \frac{π}{2} ✓$ $t h i s w o r k s e v e n w h e n w e {d o n}^{'} t k n o w$ $h o w t o c a l c u l a t e t h e c o n c r e t e i n t e g r a l s .$ $f o r t h e s a m e r e a s o n w e c a n g e t e . g .$ $\int_{0}^{\frac{π}{2}} \sin^{100} x d x + \int_{0}^{1} \sin^{- 1} (\sqrt[100]{x}) d x = \frac{π}{2}$
	Commented by mr W last updated on 29/Dec/21

	Commented by mr W last updated on 29/Dec/21

	$g e o m e t r i c a l l y \int_{0}^{a} f (x) d x m e a n s t h e$ $a r e a u n d e r t h e c u r v e y = f (x) . i . e .$ $\int_{0}^{a} y d x = \int_{0}^{a} f (x) d x = A_{1}$ $w e s e e A_{1} = a b - A_{2} w h i l e A_{2} i s t h e$ $a r e a u n d e r t h e c u r v e x = f^{- 1} (y), i . e .$ $A_{2} = \int_{0}^{b} x d y = \int_{0}^{b} f^{- 1} (y) d y .$ $t h e r e f o r e w e c a n g e t$ $\int_{0}^{a} f (x) d x = a b - \int_{0}^{b} f^{- 1} (y) d y$
	Commented by Tawa11 last updated on 30/Dec/21

	$Wow great sir .$
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