The process and answer of definite integral for X and ∫ xsin [n (A-X)] DX integral on [0, a]

The process and answer of definite integral for X and ∫ xsin [n (A-X)] DX integral on [0, a]

[Na sin (NA)] divided by the square of n
Factorization factor A ^ 4-13a ^ 2B ^ 2 + 36B ^ 2
Finding definite integral ∫ xsin ^ 2x DX [- π / 2, π / 2]
Because I can't play π / 2~
Hope to have the most concise answer
The (cos2x + 1) / 2DX of solution 1 should be (1-cos2x) / 2DX. The expert made a low-level mistake! Sin ^ 2x = (1-cos2x) / 2
Solution 1
∫xsin^2x dx
=∫x(cos2x+1)/2dx
=(1/4)∫xcos2xd2x
=(1/4)∫xdsin2x
=(1/4)xsin2x-(1/4)∫sin2xdx
=(1/4)xsin2x-(1/8)∫sin2xd2x
=(1/4)xsin2x+(1/8)cos2x [-π/2,π/2]
=[(1 / 4) (π... Expansion
Solution 1
∫xsin^2x dx
=∫x(cos2x+1)/2dx
=(1/4)∫xcos2xd2x
=(1/4)∫xdsin2x
=(1/4)xsin2x-(1/4)∫sin2xdx
=(1/4)xsin2x-(1/8)∫sin2xd2x
=(1/4)xsin2x+(1/8)cos2x [-π/2,π/2]
=[(1/4)(π/2)*sinπ+(1/8)cosπ]-[(1/4)(-π/2)*sin(-π)+(1/8)cos(-π)]
=0
Solution 2
Because f (x) = xsin ^ 2x is an odd function
Integral limit symmetric and origin
So integral = 0
The second solution is the simple answer
=1/2∫x(1-cos2x)dx
=1/2∫xdx-1/2∫xcos2xdx
=1/4x^2-1/4∫xd(sin2x）
=1 / 4-1 / 4xsin2x + 1 / 8cos2x + C
25A ^ 2-36b ^ 2 factorization,
Calculate the definite integral ∫ lower limit 1 / √ 2, upper limit 1 [√ (1-x ^ 2) / x ^ 2] DX =? I use the substitution method to calculate 1 - π / 4, the answer is π / 4 + √ 2 / 2, I use the substitution method to calculate x = Sint, what is the answer,
The answer of the book should be wrong. I calculate 1 - π / 4 as follows: ∫ √ (1-x & sup2;) / X & sup2; DX, let x = sin (U), DX = cos (U) x = 1 / √ 2, u = π / 4, x = 1, u = π / 2 = ∫ cos (U) √ [1-sin & sup2; (U)] / Sin & sup2; (U) Du = ∫ cos & sup2; (U) / Sin & sup2; (U) Du = ∫ cot & sup2
How can I use trigonometric conversion and your answer is the same
Factorization: 25A ^ 2-36b ^ 2-12b-1
Finding definite integral ∫ upper limit root sign 5, lower limit 1, under root sign (x's Square-1) / X DX by substitution method
-Factorization of AB (a-b) & # 178; + a (B-A) & # 178
-ab(a-b)²+a(b-a)²
=a（a-b)²-ab(a-b)²
=a(a-b)²(1-b)
=-ab(a-b)²+a(a-b)²
=-a(a-b)²(b-1)
Seeking indefinite integral ∫ [1 / (1 + x ^ 3)] DX
Factorization AB (a + b) ² - (a + b) ² + 1
cross multiplication
a(a+b) -1
X
b(a+b) -1
ab(a+b)²-(a+b)²+1
=[a(a+b) -1][b(a+b) -1]
=﹙a²＋ab－1﹚﹙b²＋ab－1﹚