As per the trigonometric substitution x = sin(0), the following triangle is obtained. cos(0) Submit = √₁-x² 1 2 Rewrite in terms of x. [ 16. 16 x arcsin(x) dx = 16 arcsin(x) The indefinite integral is as follows. (Use C for the constant of integration.) [16 16 x arcsin(x) dx = X Skip (you cannot come back) (2²) - 4(arcsi X arcsin(x) − x X )) + C

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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I need help with step 7 I did all of the others

Help me fill in the blanks of step 7 please

As per the trigonometric substitution x = sin(0), the following triangle is obtained.
cos(0) =
Rewrite in terms of x.
Submit
- X
1
2
[16
16 x arcsin(x) dx = 16 arcsin(x)(2²) – 4( arcsin(x) − x(
-
-
The indefinite integral is as follows. (Use C for the constant of integration.)
[ 16
X
16 x arcsin(x) dx =
Skip (you cannot come back)
X
)) + ²
C
Transcribed Image Text:As per the trigonometric substitution x = sin(0), the following triangle is obtained. cos(0) = Rewrite in terms of x. Submit - X 1 2 [16 16 x arcsin(x) dx = 16 arcsin(x)(2²) – 4( arcsin(x) − x( - - The indefinite integral is as follows. (Use C for the constant of integration.) [ 16 X 16 x arcsin(x) dx = Skip (you cannot come back) X )) + ² C
Step 4
Substitute the values of u and v in the integrand and use the formula for integration by parts.
[ 16.
16x arcsin(x) dx = 16 arcsin(x)
dx =
=
16 arcsin(x)
Step 6
Apply the half angle identity.
1
sin²(0)
cos(20)
2
16x arcsin(x) dx
Step 5
Let x = sin(0). Differentiate with respect to x on both sides.
( cos (8)
cos (0)
Rewrite the integral by substituting for x. (Simplify your answers completely.)
[₁
16x arcsin(x) dx 16 arcsin(x) (²) -
=
= 16
=
=
2
arcsin(x)
16
X
de
16 arcsin(x)
2
1
x² 16/2/²7 (√₁²-x²).
1
8/
Therefore, we have the following. (Enter your answer in terms of 0.)
[₁
arcsin(x) (2) - 4√(1- cos(20)) do
x²
3-√(√) -
dx
sin²(0)
1 - sin²(0)
( ) .
(2) - 8 / (sin² (0)
1(x) (2²2) – 4 (0 - ( sin (20)
Cos (0)
x²
sin² (0)
de
sin(20)
dx
cos (0)
+ C
de
Transcribed Image Text:Step 4 Substitute the values of u and v in the integrand and use the formula for integration by parts. [ 16. 16x arcsin(x) dx = 16 arcsin(x) dx = = 16 arcsin(x) Step 6 Apply the half angle identity. 1 sin²(0) cos(20) 2 16x arcsin(x) dx Step 5 Let x = sin(0). Differentiate with respect to x on both sides. ( cos (8) cos (0) Rewrite the integral by substituting for x. (Simplify your answers completely.) [₁ 16x arcsin(x) dx 16 arcsin(x) (²) - = = 16 = = 2 arcsin(x) 16 X de 16 arcsin(x) 2 1 x² 16/2/²7 (√₁²-x²). 1 8/ Therefore, we have the following. (Enter your answer in terms of 0.) [₁ arcsin(x) (2) - 4√(1- cos(20)) do x² 3-√(√) - dx sin²(0) 1 - sin²(0) ( ) . (2) - 8 / (sin² (0) 1(x) (2²2) – 4 (0 - ( sin (20) Cos (0) x² sin² (0) de sin(20) dx cos (0) + C de
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