Question

Đã trả lời bởi chuyên gia

Answer 1

$a, Đ ặ t u = \sin x = > d u = \cos x d x = > A = \int u^{2} d u = \frac{1}{3} u^{3} + C = \frac{1}{3} \sin^{3} x + C v ậ y A = \frac{1}{3} \sin^{3} x + C b, Đ ặ t \cos x = u = > d u = - \sin x d x = > B = - \int u^{3} d u = - \frac{1}{4} u^{4} + C = - \frac{1}{4} \cos^{4} x + C v ậ y B = - \frac{1}{4} \cos^{4} x + C c, C = \int \cos^{3} x d x = \int \cos^{2} x . \cos x d x = \int (1 - \sin^{2} x) \cos x d x = \int \cos x d x - \int \sin^{2} x . \cos x d x = \sin x - \frac{1}{3} \sin^{3} x + C (t h e o a t h ì \int \sin^{2} x . \cos x d x = \frac{1}{3} \sin^{3} x + C) V ậ y C = \sin x - \frac{1}{3} \sin^{3} x + C$

$d, D = \int x . e^{1 + x^{2}} d x Đ ặ t u = 1 + x^{2} = > d u = 2 x d x = > x d x = \frac{d u}{2} = > D = \frac{1}{2} \int e^{u} d u = \frac{1}{2} e^{u} + C = \frac{1}{2} e^{1 + x^{2}} + C v ậ y D = \frac{1}{2} e^{1 + x^{2}} + C e, E = \int \sin x . e^{1 + \cos x} d x Đ ặ t u = 1 + \cos x = > d u = - \sin x d x = > - d u = \sin x d x = > E = - \int e^{u} d u = - e^{u} + C = - e^{1 + \cos x} + C v ậ y E = - e^{1 + \cos x} + C$ ..

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Answer 2

$a, Đ ặ t u = \sin x = > d u = \cos x d x = > A = \int u^{2} d u = \frac{1}{3} u^{3} + C = \frac{1}{3} \sin^{3} x + C v ậ y A = \frac{1}{3} \sin^{3} x + C b, Đ ặ t \cos x = u = > d u = - \sin x d x = > B = - \int u^{3} d u = - \frac{1}{4} u^{4} + C = - \frac{1}{4} \cos^{4} x + C v ậ y B = - \frac{1}{4} \cos^{4} x + C c, C = \int \cos^{3} x d x = \int \cos^{2} x . \cos x d x = \int (1 - \sin^{2} x) \cos x d x = \int \cos x d x - \int \sin^{2} x . \cos x d x = \sin x - \frac{1}{3} \sin^{3} x + C (t h e o a t h ì \int \sin^{2} x . \cos x d x = \frac{1}{3} \sin^{3} x + C) V ậ y C = \sin x - \frac{1}{3} \sin^{3} x + C$

$d, D = \int x . e^{1 + x^{2}} d x Đ ặ t u = 1 + x^{2} = > d u = 2 x d x = > x d x = \frac{d u}{2} = > D = \frac{1}{2} \int e^{u} d u = \frac{1}{2} e^{u} + C = \frac{1}{2} e^{1 + x^{2}} + C v ậ y D = \frac{1}{2} e^{1 + x^{2}} + C e, E = \int \sin x . e^{1 + \cos x} d x Đ ặ t u = 1 + \cos x = > d u = - \sin x d x = > - d u = \sin x d x = > E = - \int e^{u} d u = - e^{u} + C = - e^{1 + \cos x} + C v ậ y E = - e^{1 + \cos x} + C$ ..

Answer 3

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