Ask for a high math question, ask for a high math question.

Molecule 2 = 2sint 2 + 2cost 2, cos2t = cost 2-sint 2The numerator is equal to sint 2 + 3 cost 2 = 1-2 cost 2The whole fraction becomes 1 (cost 2)+2, and the separate integral is equal to tant+2t+c.

I don't know if there is a miscalculation, the idea is like this.

In fact, it is still very simple, and the landlord should pay attention to it when doing trigonometric simplification:

1. Try to make a trigonometric function of the same form (including the differential part dx), and 2. Reduce the order as much as possible.

The first thing to know is dtanx 1 cosx 2, but also cosx 2 1 (1 tanx 2), cos2x 2cosx 2 1 2 (1 tanx 2 1).

Then the original form can be written as: (2 cos2t)dtant 2 2 (1 tant 2) 1 dtant.

Let tant x, then.

1＋2／（1＋x＾2）dx＝x＋2arctanx＝tant＋2t

If we choose a, the derivative condition of the function is continuous and both the left and right derivatives exist and are equal. The derivative of f(1) is obviously not equal to you 1, f(1) 1, continuous, so b 2. The derivative function of the function must be continuous, the function is not necessarily derivable if the function is continuous, and the function is not continuous and must not be derivable.

Because the integrand function is continuous in the defined domain, it is possible to find the limit first and then the integral, and the limit of x n in the integral region of x (excluding x=1) is equal to 0, and the denominator limit is unchanged, so the whole limit is 0, and the integral of 0 is still 0, although the limit of x=1 is not 0, but a single point does not change the value of the integral, so the result is still 0

According to the characteristics of differential equations, it can be seen that this problem is a homogeneous differential equation. The usual thing is to make y x as u, but the obvious feature here is that when x is 0, y 1. So you can't make y x as u.

So we can think differently and make x y as u. The dy dx on the left side of the equation can be reversed according to the derivative of the inverse function.

Find the differential equation y''=e (3x)+sinx.

Solution: y'=∫[e^(3x)+sinx]dx=(1/3)e^(3x)-cosx+c₁;

General solution y= [(1 3)e (3x)-cosx+c ]dx=(1 9)e (3x)-sinx+c x+c ;

linearly correlation, then the determinant is 0, d=a3 - 2 - 3a=0, then a= -1 or 2. Pick B

|y-4|"Obtainable because y=x 2

Then the root number is finally calculated as d = 2 - the root number.

The landlord means (-1) (n-1) divided by (2n-1) (2n+1).

1)^(n-1) / [(2n-1)(2n+1)]= 1/2 [ 1)^(n-1) / (2n-1) -1)^(n-1) / (2n+1) ]

The sum of the first half (-1) (n-1) (2n-1) is 1 + 1 - 1 3 + 1 5 - = 1 + 4

The sum of the last half (-1) (n-1) (2n+1) is -1 + 1 3 - 1 5 + = - 4

Therefore, the value in parentheses is 1 + 4 - 4) = 1 + 2, and the value of the original is 1 2 + 4

I don't know, right?

I'm just working on this!

Is it (2n-1) 2*(-1) (n-1)? If so, this is a divergent series and cannot be summed.

Note: On behalf of the power, * is a multiplier sign, please add more brackets when writing, or write a screenshot in word and send it to me, then there will be no problem with reading the question, thank you for your cooperation!

Some divergent series can also sum functions, but they diverge from and functions, the simplest example is the summation of a series of equal differences