please fast how to integrate (2x+3arcsen(x))/√(1-x^2 )?
integrate (2x+3arcsen(x))/√(1-x^2 )
Update:.∫(2x+3arcsen(x))/√(1-x^2 )dx
∫2x/√(1-x^2 ) dx +3∫(arcsen(x))/√(1-x^2 ) dx=> 3∫(arcsen(x))/√(1-x^2 ) dx
=∫〖f^' (x).[f(x)]^n 〗dx = [f(x)]^(n+1)/(n+1)+c = 3 〖arcsen(x)〗^2/2 +c
More Questions From This User See All
Answers & Comments
The integral may be written as:
2 ∫ x / sqrt(1-x^2) dx + 3 ∫ arcsin(x) /sqrt(1-x^2) dx
2 ∫ x / sqrt(1-x^2) dx
Let u=1-x^2
du = -2 x dx
x dx = (-1/2) du
2 ∫ x / sqrt(1-x^2) dx = -2(1/2) ∫ du/sqrt(u) = - ∫ u^(-1/2) du = - u^(-1/2+1)/(-1/2+1)
= -2 u^(1/2)
= -2 sqrt(1-x^2) ------ (1)
3 ∫ arcsin(x) /sqrt(1-x^2) dx
Let u = arcsin(x)
du = 1/sqrt(1-x^2) du
3 ∫ arcsin(x) /sqrt(1-x^2) dx = 3 ∫ u du = (3/2) u^2
= (3/2) (arcsin (x) )^2 ------ (2)
(1)+(2)
= -2 sqrt(1-x^2) + (3/2) (arcsin (x) )^2 + C
1/√(1-x^2 ) is the derivative of arcsen(x). Integrate by parts.