What is the original function of the fifth power of sin?

What is the original function of the fifth power of sin?

∫(sinx)^5dx
=-∫(sinx)^4dcosx
=-∫[1-(cosx)^2]^2dcosx
=-∫[(cosx)^4-2(cosx)^2+1]dcosx
=-(cosx)^5/5+2(cosx)^3/3-cosx+C
C is the integral constant
The algebraic identity (2a-b) (a + 2b) = the square of 2A + the square of 3ab-2b
On the two roots of the unary linear equation x & # 178; + 3x + M-1 = 0 of X are x1, x2
(1) Find the value range of M (2) if 2 (x1 + x2) + x1x2 + 10 = 0
It is known that the quadratic equation x & # 178; + MX + n-1 = 0 with respect to X. if the equation has two equal real roots, find the value of M & # 178; - 2n + 2 / n-1
1、
△>=0
So 9-4 (m-1) > = 0
M-1
What is the period of the function y = cos sixth power a + sin sixth power a?
Cos sixth power a + sin sixth power a
=(cos^2a)^3+(sin^2a)^3
=(cos^2a+sin^2a)(cos^4a-cos^2asin^2a+sin^4a)
=(cos^2a+sin^2a)^2-3cos^2asin^2a
=1-3/4(sin2a)^2
=1-3/4(1-cos4a)/2
=1-3/8+3/8cos4a
=5/8+3cos4a/8
So the period is 2 π / 4 = π / 2
(5a squared - 3AB + B squared) - (2a squared + 3ab-2b squared) where a squared + B squared = 10 AB = 3
If we remove the brackets, we will get = 5A & # 178; - 3AB + B & # 178; - 2A & # 178; - 3AB + 2B & # 178; simplify to = 3A & # 178; - 6ab + 3B & # 178; 3 (A & # 178; + B & # 178;) = 3 × 10
6ab=6×3 ∴=30-18=12
I've been fighting for a long time,
It is known that (M 2-1) x 2 + (M + 1) x + 1 = 0 is a linear equation of one variable about X, and the value of M is obtained
∵ (M2-1) x2 + (M + 1) x + 1 = 0 is a linear equation with one variable about X, where ∵ M2 − 1 = 0m + 1 ≠ 0, the solution is m = 1
What is the original function of one fourth power of (sin a)?
First, we use the first kind of substitution method
(1/[sin(x)]^4)*dx=-(1/[sin(x)]^2)*dcot(x)=-(1+[cot(x)]^2)*dcot(x)=-(cot(x)+[cot(x)]^3/3)
Let 2A power = 5B power = m, and 1 / A + 1 / b = 2
Let's make it easy!
2^a=5^b=m,
m^(1/a)=2,m^(1/b)=5
m^(1/a+1/b)=10
M ^ 2 = 10 (known by, M > 0)
So m = √ 10
2^a=5^b=m
alg2=blg5
a=blg5/lg2
1/a+1/b=2
a. B is not 0
a+b=2ab
blg5/lg2+b=2*(blg5/lg2)*b
lg5/lg2+1=2*(blg5/lg2)
lg5+lg2=2blg5
lg10=lg5^2b
5^2b=10
5^b=√10
m=√10
In fact, I also want to ask this question, what school are you from??? Follow up: Lanzhou 27
If the equation (M + 2) x & # 178; + 4mx-4m = 0 about X is a linear equation with one variable, then the solution of the equation is () a - 1 B 1 C 0 D
B
You must be right
How to draw the golden section of a line with ruler?
The method of drawing with ruler and gauge needs to be described step by step, i.e. to cut the compass and use ruler
Please answer how to draw a vertical line with ruler and how to make one line become half of another line
Let a given line segment be AB, and BC ⊥ AB be BC ⊥ AB through B, and BC = AB / 2; 2. Connect AC; 3. Take C as the center of the circle, CB as the radius of the arc, intersect AC at d; 4. Take a as the center of the circle, ad as the radius of the arc, intersect AB at P, then point P is the golden section point of ab
That's it!