The calculator will accept any of the following expressions :

**Straight lines:**(3x - 2)**Polynomials:**(x^3 + 3x^2 - 5x + 2)- Any of the
**trigonometric functions:**sin(x), cos(x/2), tan(2x), csc(3x), sec(x/4), cot(x) - The
**inverse trigonometric functions:**arcsin(x), arccos(x), arctan(x), arccsc(x), arcsec(x), arccot(x) **Exponential**(e^x) and logarithm (ln(x) for natural log and log(x) for log base 10)**Absolute value**: use "abs" like this: abs(x)- The
**hyperbolic functions and their inverses:**sinh(x), cosh(x), tanh(x), arcsinh(x), arccosh(x), arctanh(x)

**ceiling:**ceil(x) and**round**: round(x)**square root:**sqrt(x)

You can also use any combinations of the above, like "ln(abs(x))".

In mathematics, integration is the act of calculating an integral. It is also one of the two branches of infinitesimal calculus, also called integral calculus, the other being differential calculus. Since the operations of measuring quantities (length of a curve, area, volume, flow, etc.) and calculating probabilities are often subjected to calculations of integrals, integration is a fundamental scientific tool. This is the reason why integration is often tackled from secondary education.

The solution to the integral of sin^2(x) requires you to recall principles of both trigonometry and calculus.

For sin2(X), we will use the cos double angle formula:

cos(2X) = 1 - 2sin²(X)

The above formula can be rearranged to make sin²(X) the subject:

sin²(X) = 1/2(1 - cos(2X))

You can now rewrite the integration:

∫sin²(X)dX = ∫1/2(1 - cos(2X))dX

Because 1/2 is a constant, we can remove it from the integration to make the calculation simpler. We are now integrating:

1/2 x ∫(1 - cos(2X)) dX = 1/2 x (X - 1/2sin(2X)) + C

It is very important that as this is not a definite integral, we must add the constant C at the end of the integration.

Simplifying the above equation gives us a final answer:

∫sin²(X) dX = 1/2X - 1/4sin(2X) + C

Integration by Parts.

∫ln(x) dx

set u = ln(x), dv = dx then we find du = (1/x) dx, v = x

substitute

∫ ln(x) dx = ∫ u dv

and use integration by parts

= uv - ∫ v du

substitute u=ln(x), v=x, and du=(1/x)dx

= ln(x) x - ∫ x (1/x) dx

= ln(x) x - ∫ dx

= ln(x) x - x + C

= x ln(x) - x + C.

Since the integral is defined by taking the area under the curve, an integral can be taken from any continuous function, because the area can be found.However, it is not always possible to find the indefinite integral of a function by basic integration techniques.

The dx represents an infinitesimal variation of the variable x. The integral of f is therefore the sum of the basic geometrical elements dx and of height f (x) between the integration limits. It is a continuous sum, i.e. the limit of the discrete sum (sigma) when dx approaches 0.

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