Grade 12Integral Calculus

integration of sinx/x with respect to x

hitesh jaiswal

14 Years agoGrade 12

2 Answers

bhanuveer danduboyina

14 Years ago

The fact of the matter is, this integral is one that cannot be expressed in terms of elementary functions. There's no way we can solve this using the methods we know; we cannot use integration by parts, partial fractions, substitution, trigonometric substitution, etc to solve this.

We can, however, approximate the integral through a power series. sin(x) has its own power series, so all we need to do is divide each term of the series by x (this represents (1/x)sin(x), or sin(x)/x) and then integrate thereafter.

como: Your proposed solution doesn't work, and here's why. Let

f(x) = (1/x)cos x - (1/x²)sinx + C

To make it easier to differentiate, factor (1/x).

f(x) = (1/x) [cos(x) - (1/x)sin(x)] + C

Differentiate using the product rule, noting that d/dx (1/x) = -1/x^2 gives us

f'(x) = (-1/x^2) [cos(x) - (1/x)sin(x)] + (1/x) [-sin(x) - [(-1/x^2)sin(x) + (1/x)cos(x)] ]

f'(x) = -cos(x)/x^2 + sin(x)/x^3 - sin(x)/x + sin(x)/x^2 - cos(x)/x

And as you can see, it looks nothing like sin(x)/x.

PLEASE APPROVE MY ANSWER!!

Rakib

6 Years ago

Alternatively, you could have used the ordinary integration by parts: ∫ sin x x d x = − ∫ 1 x d ( cos x ) = − cos x x − ∫ cos x x 2 d x = − cos x x − ∫ 1 x 2 d ( sin x ) = − cos x x − sin x x 2 − 2 ∫ sin x x 3 d x ∫sin⁡xxdx=−∫1xd(cos⁡x)=−cos⁡xx−∫cos⁡xx2dx=−cos⁡xx−∫1x2d(sin⁡x)=−cos⁡xx−sin⁡xx2−2∫sin⁡xx3dx You may notice a recursion for the integrals: I n ≡ ∫ sin x x 2 n + 1 d x In≡∫sin⁡xx2n+1dx By similar double integration by parts, you get: I n = − cos x x 2 n + 1 − ( 2 n + 1 ) sin x x 2 n + 2 − ( 2 n + 1 ) ( 2 n + 2 ) I n + 1 In=−cos⁡xx2n+1−(2n+1)sin⁡xx2n+2−(2n+1)(2n+2)In+1 As you can see, this procedure goes on indefinitely and you do not get a closed form.