Integral involving parts and trig sub

[Punkyc7]

integral of xarcsinx dx

integration by parts and a trig sub


by parts I get (x^2)/2 arcsinx - integral (x^2)/sqrt(1-x^2)

after that trig sub and i get

integral of sin^2

then I used the double angle identity

so

integral of 1/2 - the integral of (cos2)/2

so i get 1/2 theta - 1/4 sincos

then i plug bach the values of sin, cos and theta and I get

x^2)/2 arcsinx-1/2 arcsinx +1/4 xsqrt(1-x^2)


the answer that is given is x^2)/2 arcsinx-1/4 arcsinx +1/4 xsqrt(1-x^2)

my question is how do you get the 1/4 infront of the arcsinx

 

[Mark44]

integral of xarcsinx dx

integration by parts and a trig sub


by parts I get (x^2)/2 arcsinx - integral (x^2)/sqrt(1-x^2)

That's not what I get. You left of the factor of 1/2 in the integral. You have also omitted dx, which will come around to bite you later.

after that trig sub and i get

integral of sin^2

That's not what I get. Before the trig substitution, the integral is
$$(1/2)\int \frac{x^2~dx}{\sqrt{1 - x^2}}$$

You used the right trig substitution, but you didn't substitute for everything. It looks like all you did was substitute for x², and you didn't substitute for either dx or sqrt(1 - x²).

then I used the double angle identity

so

integral of 1/2 - the integral of (cos2)/2

so i get 1/2 theta - 1/4 sincos

then i plug bach the values of sin, cos and theta and I get

x^2)/2 arcsinx-1/2 arcsinx +1/4 xsqrt(1-x^2)


the answer that is given is x^2)/2 arcsinx-1/4 arcsinx +1/4 xsqrt(1-x^2)

my question is how do you get the 1/4 infront of the arcsinx