Solving indefinite integral ∫ sin2x × 3 ^ cosx ^ 2 × quadratic root sign (1 + 3 ^ cosx ^ 2 ×) DX

Solving indefinite integral ∫ sin2x × 3 ^ cosx ^ 2 × quadratic root sign (1 + 3 ^ cosx ^ 2 ×) DX

Let u = 3 ^ cos & # 178; X, Du = 3 ^ cos & # 178; X * Ln3 * 2cosx * - SiNx DX
= -3^cos²x * ln3 * sin(2x) dx
∫ sin(2x) * 3^cos²x * √(1 + 3^cos²x) dx
= -1/ln3 ∫ √(1 + u) du
= -1/ln3 ∫ √(1 + u) d(1 + u)
= -1/ln3 * (2/3)(1 + u)^(3/2) + C
= -1/ln3 * (2/3)(1 + 3^cos²x)^(3/2) + C
= -[2(1 + 3^cos²x)^(3/2)]/(3ln3) + C

Vector a = (sinx-1), vector b = (1, cosx), function f (x) = vector a * vector B, find the range

If f (x) = SiNx cosx = √ 2 [(√ 2 / 2) SiNx - (√ 2 / 2) cosx] = √ 2Sin (x - π / 4), then: F (x) ∈ [- √ 2, √ 2]

It is known that the image of the first function y = 2x + 1 passes through point a (a, - 3), and the image vertex of the second function y = xsquare - (M + 1) x + m is d
(1) Verification: the image of this quadratic function must intersect with the X axis
(2) When the image of a quadratic function passes through point a, the analytic expression of the quadratic function is obtained

(1) You can "decompose" the quadratic function first! Y = (x-1) (x-m)... Right! Obviously, if y = 0, x = 1 or x = m, the solution of x = 1 is inevitable, that is, there is an intersection between the quadratic function and the X axis! (2) first, take point a into the primary function. 2 * a + 1 = - 3. There is a = - 2, that is, point a is (- 2, - 3). From the title, we can see that the quadratic function passes through point a! Take point a into account