The following indefinite integrals are obtained by the first type of substitution method (approximate differentiation method) or the second type of substitution method: 1）∫dx/[x*(x^6+4)]; (1/24)*ln[x^6/(x^6+4)]+c; 2) ∫cosxcos(x/2)dx ; : (1/3)sin(3x/2)+sin(x/2)+c; 3) ∫tan^3secxdx; : (1/3)(secx)^3-secx+c; 4 ∫dx/[x√(x^2-1)]; : arccos(1/x)+c; The answer has been given, please give the process. Do not require all answers, solve one by one

The following indefinite integrals are obtained by the first type of substitution method (approximate differentiation method) or the second type of substitution method: 1）∫dx/[x*(x^6+4)]; (1/24)*ln[x^6/(x^6+4)]+c; 2) ∫cosxcos(x/2)dx ; : (1/3)sin(3x/2)+sin(x/2)+c; 3) ∫tan^3secxdx; : (1/3)(secx)^3-secx+c; 4 ∫dx/[x√(x^2-1)]; : arccos(1/x)+c; The answer has been given, please give the process. Do not require all answers, solve one by one

The answer to the second question should be (- 4 / 3) {[sin (x / 2)] to the third power} + 2Sin (x / 2). You can lead it yourself
Expand cosx into one with sin square term, and then put it in DX to become DCOS (x / 2). The latter should know it by themselves. Think of this first, it's a headache

Question of inverse trigonometric function I asked for COS in a topic α= Negative root 6 / 4, last α= Arccos root 6 / 4. Another cos β= Negative 1 / 4, result β= arccos1/4 [the minus sign is gone. Is this for god horse?]

To see the value range of the inverse triangle, arccosx is between 0 and PI;
If it is an acute angle, the independent variable is positive;
If it is an obtuse angle, the independent variable is negative;
Another: you find cos α= The negative root sign 6 / 4 is an obvious error: cos α The absolute value of is > 1

Inverse trigonometric function problem Is arctan (√ 2x + 1) + arctan (√ 2x-1) equal to arctan √ 2 / 2 (x-1 / x)? How?

Let a = arctan (√ 2x + 1)
b=arctan(√2x-1)
c=arctan√2/2（x-1/x）
Then √ 2x + 1 = Tana
√2x-1=tanb
√2/2（x-1/x)=tanc;
Title: a + B = C;
To a + B = C
Tan (a + b) = Tanc is required;
Expand and bring in X~

What is the image of the inverse trigonometric function?

Y = arcsin (x), domain [- 1,1]  , The range [- π / 2, π / 2] image is marked with red lines;     Y = arccos (x), domain [- 1,1]  ,  Value range [0, π], the image uses blue lines;     Y = arctan (x), definition domain (- ∞, + ∞), value domain (- π / 2, π / 2), and the image uses green lines;     sin(arcsin   x) = x, definition field [- 1,1], value field   [-1,1]   arcsin(-x)=-arcsinx     The proof method is as follows: let arcsin (x) = y, then sin (y) = X  , Substitute these two formulas into the above formula     Several others can be obtained by similar methods     cos(arccos   x)=x,   arccos(-x)=π-arccos   x     tan(arctan   x)=x,   arctan(-x)=-arctanx

What is an inverse trigonometric function? What is its basic definition and algorithm?

Inverse trigonometric function is a mathematical term. It can not be narrowly understood as the inverse function of trigonometric function. It is a multivalued function. It is the general name of inverse sine arcsin x, inverse cosine arccos x, arctan X and inverse cotangent arccot X. each of these functions represents the angle whose sine, cosine, tangent and cotangent are X

What's the usage? Why use it? Such as the title!

It is to know the value of a trigonometric function and find the size of the angle in turn. This is the purpose. It is usually pressed by the computer, such as SIN-1, written by hand in arcsin (xxx). It is used to find the size of the angle