Detailed explanation of a definite integral problem about inverse trigonometric functions

A question about the definite integral of an inverse trigonometric function is troublesome with detailed procedures

∫ (arcsinx)² dx

= x(arcsinx)² - ∫ x d(arcsinx)²

= x(arcsinx)² - ∫ x • 2(arcsinx) • 1/√(1 - x²) • dx

= x(arcsinx)² - 2∫ x(arcsinx)/√(1 - x²) dx

= x(arcsinx)² - 2∫ arcsinx d[-√(1 - x²)]

= x(arcsinx)² 2(arcsinx)√(1 - x²) - 2∫ √(1 - x²) d(arcsinx)

= x(arcsinx)² 2(arcsinx)√ (1 - x²) - 2∫ √ (1 - x²)/√ (1 - x²) dx

= x(arcsinx)² 2(arcsinx)√(1 - x²) - 2x C

This is an indefinite integral

Just substitute the fixed points

Original function of inverse trigonometric function

Use the integral method to get:

I = ∫ arcsinx dx = x arcsinx - ∫ [x/√（1-x^2）] dx

= x arcsinx (1/2) ∫ [1/√（1-x^2）] d(1-x^2) = x arcsinx √（1-x^2） C

I = ∫ arccosx dx = x arccosx ∫ [x/√（1-x^2）] dx

= x arccosx - (1/2) ∫ [1/√（1-x^2）] d(1-x^2) = x arccosx - √（1-x^2） C

I = ∫ arctanx dx = x arctanx - ∫ [x/(1 x^2)] dx

= x arctanx - (1/2) ∫ [1/(1 x^2)] d(1 x^2) = x arctanx - (1/2)ln(1 x^2) C

It is the general name of arcsine arcsin x, arccosine arccos x, arctangent arctan x, arccotangent arccot ​​x, arcsec arcsec x, arccscc x, each of which represents its arcsine, arccosine, Inverse tangent, inverse cotangent, inverse secant, and inverse cosecant are the angles of x.

Extended information:

It is best for the function to be continuous in this interval (the reason why it is said to be best here is because the inverse secant and inverse cosecant functions are sharp); in order to facilitate research, it is often necessary to select the interval from 0 to π/ 2 corners.

The function value domain on the determined interval should be the same as the domain of the entire function. The inverse trigonometric function determined in this way is single-valued. In order to distinguish it from the multi-valued inverse trigonometric function above, the A in Arc is often changed to a in notation. For example, the single-valued inverse sine function is recorded as arcsin x.

To limit the inverse trigonometric function to a single-valued function, limit the value y of the inverse sine function to -π/2≤y≤π/2, and use y as the principal value of the inverse sine function, recorded as y=arcsin x; Correspondingly, the principal value of the inverse cosine function y=arccos x is limited to 0≤y≤π; the principal value of the arctangent function y=arctan x is limited to -π/2

Reference source: Encyclopedia - Inverse Trigonometric Functions

How to prove the indefinite integral of an inverse trigonometric function

. If the integral interval is symmetric, first check whether there is an odd function in the formula. For example, the square expansion of this question is: 1 2x(1-x^2)^1/2. Note that 2x(1-x^2)^ 1/2 is an odd function, so its integral in the symmetric interval is 0, leaving only "1", so the result is 2

2. When arctan, ln and the like appear, you must find a way to make derivatives of them, x*arctanx. If you want to make derivatives of arctanx, you must use integrals by parts:

Put x at the back, the original integral formula becomes: 1/2arctanx d(x^2), the integral formula of the second half of the integral by parts is (x^2)/(1 x^2), this should It will accumulate, the key is to know how to guide arctan

The result of this question is: 1/2(x^2*arctanx - x arctanx C)

As long as you do more questions here, you will get the idea. The real difficulty lies in the multiple integrals and surface curve integrals, which can be said to be abnormal

Derivation of integral formula by parts

The integral-by-parts formula is a very important formula. With it, the formula can be used to quickly solve some integral problems. At the same time, the answer can also be solved when some integrand functions cannot directly find the original function.

Extended information:

1. The integral method by parts is an important and basic method for calculating integrals in calculus.

2. It is derived from the multiplication rule of differential calculus and the fundamental theorem of calculus. Its main principle is to transform the integral form that is not easy to produce direct results into an equivalent integral form that is easy to produce results.

3. According to the basic function types that make up the integrand, the order of commonly used integrals by parts is organized into a formula: "Anti-power refers to three". They respectively refer to five types of basic functions: inverse trigonometric functions, logarithmic functions, power functions, exponential functions, and integrals of trigonometric functions.

4. Formula (1) of indefinite integral, ∫ a dx = ax C, a and C are constants

(2), ∫ x^a dx = [x^(a 1)]/(a 1) C, where a is a constant and a ≠ -1

(3), ∫ 1/x dx = ln|x|

(4), ∫ a^x dx = (1/lna)a^x C, where a > 0 and a ≠

(5), ∫ e^x dx = e^x C

(6), ∫ cosx dx = sinx

(7), ∫ sinx dx = - cosx C

(8), ∫ cotx dx = ln|sinx| C = - ln|cscx| C

5. Method of indefinite integral:

The first type of substitution is actually a kind of patchwork, using f'(x)dx=df(x); and the rest of the previous ones are just functions about f(x), and then look at f(x) To achieve the final result as a whole.

Integral by parts, there are only a few fixed types, which are nothing more than trigonometric functions multiplied by x, or exponential functions or logarithmic functions multiplied by an x. The memory method is to use the f mentioned above '(x)dx=df(x) deform, and then use the formula ∫xdf(x)=f(x)x-∫f(x)dx. Of course, x can be replaced by other g(x).

Reference: Encyclopedia: Integration by Parts

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