how to integrate (x+1)sin^-1(x+2) dx?

I give just a hint; you may try to solve yourself.

1) Let sin⁻¹(x+2) = t; ==> x+2 = sin(t)

Differentiating both sides, dx = cos(t) dt

2) x+1 = sin(t) - 1

Thus the given one is: ∫[sin(t) - 1]*t*cos(t) dt

==> = ∫t*sin(t)*cos(t) dt - ∫t*cos(t) dt

==> = (1/2)∫t*sin(2t) dt - ∫t*cos(t) dt

3) Each of the above, you may integrate using parts rule of integration and finally substitute t = sin⁻¹(x+2), which will give you the answer.

u = x^2 + a million u - a million = x^2 sqrt(u - a million) = |x| If we take an essential from x = a million to x = 2, or certainly as you reported, from x = 2 to x = a million, in both case x is continuously constructive as a thanks to drop absolutely the price operation. i will assume that you meant a million to be the decrease certain. If no longer, purely replace the connect up our very last answer. x = sqrt(u - a million) dx = a million /(2*sqrt(u - a million)) du Now we may be able to rewrite our essential as follows, replacing our limits of integration to be in words of u particularly of x ?[from u=2 to u=5] (sqrt(u - a million)*u^3) / (2sqrt(u - a million)) du = ?[2:5] (u^3)/2 du = (a million/8)u^4 ]{2:5} we may be able to now enter rapidly = (a million/8)[5^4 - 2^4] = 609/8 or go back to a form in words of x =(a million/8)(x^2 + a million)^4]{a million:2} =(a million/8)[5^4 - 2^4] = 609/8 --charlie

Page 1
http://www.flickr.com/photos/61664301@N06/6551552151/

Page 2 (continues)
http://www.flickr.com/photos/61664301@N06/6551564683/

Page 3 (last page)
http://www.flickr.com/photos/61664301@N06/6551576083/

(x+1) sin^-1(x+2) was written as (x+2-1) sin^-1(x+2)
= Integral (x+2) sin^-1(x+2) - Integral sin^-1(x+2)

Integral sin^-1(x+2) completed in (3)
Integral (x+2) sin^-1(x+2) completed in (6)
Answer = (6)- (3)

First, let w = x+2.
So, x = w-2 and dx = dw.

Therefore, ∫ (x+1) arcsin(x+2) dx
= ∫ ((w-2) + 1) arcsin w * dw
= ∫ (w-1) arcsin w dw.

Next, let w = sin z, dw = cos z dw.
==> ∫ (sin z - 1) z * (cos z dz)
= ∫ z [(1/2) sin(2z) - cos z] dz.

Now, use integration by parts with
u = z, dv = [(1/2) sin(2z) - cos z] dz
du = dz, v = (-1/4) cos(2z) - sin z.

So, ∫ z [(1/2) sin(2z) - cos z] dz
= z [(-1/4) cos(2z) - sin z] - ∫ [(-1/4) cos(2z) - sin z] dz
= z [(-1/4) cos(2z) - sin z] - [(-1/8) sin(2z) + cos z] + C
= z [(-1/4) (cos^2(z) - sin^2(z)) - sin z] - [(-1/4) sin z cos z + cos z] + C
= arcsin w * [(-1/4) ((1 - w^2) - w^2) - w] - [(-1/4) w √(1 - w^2) + √(1 - w^2)] + C,
via w = sin z and √(1 - w^2) = cos z

= (1/4) [(1 - 2w^2 - 4w) arcsin w + (w - 4) √(1 - w^2)] + C
= (1/4) [(1 - 2(x+2)^2 - 4(x+2)) arcsin(x+2) + ((x+2) - 4) √(1 - (x+2)^2)] + C.

I hope this helps!