Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

integration of sinx/x with respect to x

integration of sinx/x with respect to x


 

Grade:12

2 Answers

bhanuveer danduboyina
95 Points
9 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
15 Points
one year 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.

 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free