A quick question, what is the integral of: a sin(x) dx? I mean like this -> ∫asin(x) dx.?

May 2021 6 81 Report

A quick question, what is the integral of: a sin(x) dx? I mean like this -> ∫asin(x) dx.?

Note: 'a' sometimes depend of 'x', and other times not. I mean like this a=x.

Then it can be like this -> ∫xsin(x)

Also not only it have to be 'sin' it also can be 'cos' or any trigonometric function. It can be like this -> ∫acos(x) dx.

Because if i do it by integration by parts or chain rule, then the output is infinite or periodic, i hope you understand what i mean.

Update:

Because since ∫u(t) dt = (1/2)t²u(t)

Science & Mathematics / Mathematics

∫ xsin(x) dx

u = x du = dx

dv = sin(x) dx v = -cos(x)

∫ x sin(x) dx = -xcos(x) +∫ cos(x) dx

= -xcos(x) + sin(x)

differentiate to check

-cos(x) + xsin(x) + cos(x) = xsin(x)

So this can be done by parts.

∫ e^x sin(x) dx

use integration by parts twice and you end up with an equation that has the target integral on both sides. Solve for the target integral as an unknown.

It looks like you are thinking about something called an integral transform.

∫ f(x) g(x) dx, for a fixed g(x), as a function of f(x). This is covered in advanced calculus or differential equations.

However it is also possible that you are not asking such a sophisticated question and instead need to learn more calculus.

Chameleojack

1) a is a constant

∫a*sinx dx = a∫sinx dx = -a*cosx + C,

per standard integration

2) a is not a constant, that is, a is a function of x. Then that depends on what the function of x is.

∫[f(x)]*sinx dx = ?

If f(x) is a polynomial, integration by parts will work just fine.

∫u dv = uv - ∫v du

Since sin(x) is a function whose derivatives repeat in cycles, and derivatives of polynomials eventually go to zero, let u = f(x) and dv = sin(x). The number of iterations will depend on the degree of the polynomial.

But that is only one case.

Tophe

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