Solving indefinite integral: ∫ DX / [(SiNx) ^ 2cosx]

Solving indefinite integral: ∫ DX / [(SiNx) ^ 2cosx]

The original formula = ∫ [(SiNx) ^ 2 + (cosx) ^ 2] DX / (SiNx) ^ 2cosx = ∫ DX / cosx + ∫ cosxdx / (SiNx) ^ 2
The first = ∫ cosxdx / (COX) ^ 2 = ∫ D (SiNx) / 1 - (SiNx) ^ 2
Second = ∫ D (SiNx) / (SiNx) ^ 2
Factorization 121 (a ^ 2-2ab + B ^ 2) - 169 (a ^ 2 + 2ab-b ^ 2)
121(a^2-2ab+b^2)-169(a^2+2ab-b^2)
=[11(a-b)]^2-[13(a+b)]^2
=(11a-11b+13a+13b)(11a-11b-13a-13b)
=(24a+2b)(-2a-24b)
=-4(12a+b)(a+12b)
121(a^2-2ab+b^2)-169(a^2+2ab+b^2)
=【11(a-b)】²-【13(a+b)】²
=（11a-11b+13a+13b）（11a-11b-13a-13b）
=(24a+2b)(-2a-24b)
=-4(12a+b)(a+12b)
121(a^2-2ab+b^2)-169(a^2+2ab+b^2)
=121(a-b)²-169(a+b)²
=[11(a-b)-13(a+b)][11(a-b)+13(a+b)]
=(11a-11b-13a-13b)(11a-11b+13a+13b)
=(-2a-24b)(24a+2b)
=-4(a+12b)(12a+b)
121(a^2-2ab+b^2)-169(a^2+2ab+b^2)
=11^2(a-b)^2-13^2(a+b)^2
=(11a-11b+13a+13b)(11a-11b-13a-13b)
=(24a+2b)(-2a-24b)
=-4(12a+b)(a+12b)
Finding the indefinite integral of 1 / (SiNx + 2cosx + 3)
Consider the half angle formula, let t = Tan (x / 2), use this formula to replace, SiNx = 2T / (1 + T ^ 2), cosx = (1-T ^ 2) / (1 + T ^ 2), DX = 2 / (1 + T ^ 2); the original formula = 1 / ((T + 1) ^ + 4), use the derivative property of arctanx, the indefinite integral result is 0.5arctan ((T + 1) / 2) + C, take t = Tan (x / 2) in
Factorization of 25 (a + b) & # 178; - 9 (a-b) & # 178
25(a+b)²-9(a-b)²
=[5(a+b)+3(a-b)][(5(a+b)-3(a-b)]
=(8a+2b)(2a+8b)
From the meaning of the title, we can get [5 * (a + b) + 3 * (a-b)] * [5 * (a + b) - 3 * (a-b)] = (8a + 2b) * (2a + 8b)
Find the third root of indefinite integral (SiNx + cosx) divided by (SiNx cos x)
Using the method of approximate differentiation
∫(sinx+cosx) * (sinx-cosx)^(-1/3) dx
It is observed that D (SiNx cosx) = (cosx + SiNx) DX
So the original formula = ∫ (SiNx cosx) ^ (- 1 / 3) d (SiNx cosx)
=(3/2) * (sinx-cosx)^(2/3) + C
finish
Factorization and cross multiplication
48+13(a-b)-(a-b)²
48+13(a-b)-(a-b)²
=-[(a-b)²-13(a-b)-48]
=-(a-b-16)(a-b+3)
Finding x ^ 2 + (SiNx) ^ 2 / x ^ 2 * (SiNx) ^ 2 indefinite integral process
∫ (x²+sin²x)/(x²sin²x) dx=∫ [x²/(x²sin²x)+sin²x/(x²sin²x)] dx=∫ (1/sin²x+1/x²) dx=∫ [csc²x+x^(-2)] dx=-cotx+x^(-2+1)/(-2+1)+C=-cotx...
Let's split the items first
The original formula = ∫ CSC ^ 2xdx + ∫ [x ^ (- 2)] DX
=-cotx-x^(-1)+C
Several problems in factorization of cross multiplication
1.x^2+2x-8
2.x^2+3x-10
3.x^2-x-20
4.x^2+x-6
5.2x^2+5x-3
6.6x^2+4x-2
7.x^2-2x-3
8.x^2+6x+8
9.x^2-x-12
10.x^2-7x+10
11.6x^2+x+2
12.4x^2+4x-3
1.x^2+2x-8 （x-2）(x+4)-2*4=-8;-2+4=22.x^2+3x-10（x+5）(x-2)5*(-2)=-10;-2+5=33.x^2-x-20 （x-5）(x+4)-5*4=-20;-5+4=-14.x^2+x-6 （x-2）(x+3)-2*3=-6;-2+3=15.2x^2+5x-3 （2x-1）(x+3)-1*3=-3;2*3+(-1)*1=56.6...
If we know SiNx = 2cosx, we can find six trigonal function values of angle X?
Given SiNx = 2cosx, we can find six 3-angle function values of angle X? ∵ SiNx = 2cosx ＾ TaNx = 2, Cotx = 1 / TaNx = 1 / 2 ∵ SiNx = 2cosx ＾ X in the first or third quadrant cosx = ± √ {1 / [1 + (TaNx) ＾]} = ± (√ 5) / 5 secx = ± √ 5, SiNx = ± √ [1 - (cosx) ＾] = ± 2 (√ 5) / 5 CSCX =
sinx/cosx=2
So TaNx = 2
cotx=1/tanx=1/2
(sinx)^2+(cosx)^2=1
sinx=2cosx
So 5 (cosx) ^ 2 = 1
(cosx)^2=1/5
cosx=±√5/5
sinx=2cosx=±2√5/5
secx=1/cosx=±√5
cscx=1/sinx=±√5/2
Factorization problem
(1-1/2~)(1-1/3~)...(1-1/10~)
How to calculate it
Where ~ is the square. For example, 1 / 2 ~ is the square of two
=(1 + 1 / 2) (1-1 / 2) (1 + 1 / 3) (1-1 / 3)... (1 + 1 / 10) (1-1 / 10) = 3 / 2 * 1 / 2 * 4 / 3 * 2 / 3. * 11 / 10 * 9 / 10. Then you multiply 1, 3, 5... Singular items together on the draft paper, and you will find the rule: the numerator of the former item and the denominator of the latter item can be roughly divided. Similarly, put 2, 4, 6... Even items together
Can be equal to (1-1 / 2) (1 + 1 / 2) (1-1 / 3) (1 + 1 / 3) (1-1/10)(1+1/10)=11/20