Cho $\Large a$ là số thực dương. Tính $\Large I = \int_{0}^{a}\sin^{20

Cho $\Large a$ là số thực dương. Tính $\Large I = \int_{0}^{a}\sin^{2016}x . \cos (2018x)dx$ bằng:

• A.

  A. $\Large I = \dfrac{\cos^{2017}a. \sin 2017a}{2016}.$

• B.

  B. $\Large I = \dfrac{\sin^{2017}a. \cos 2017a}{2017}.$

• C.

  C. $\Large I = \dfrac{\sin^{2017}a. \cos 2017a}{2016}.$

• D.

  D. $\Large I = \dfrac{\cos^{2017}a. \cos 2017a}{2017}.$

Chọn B

Ta có:

$\Large I = \int_{0}^{1}\sin^{2016}x . \cos (2018x)dx$

$\Large =\int_{0}^{a} \sin^{2016}x.\left [ \cos(2017x).\cos x - \sin(2017x).\sin x \right ]dx$

$\Large =\int_{0}^{a} \sin^{2016}x.\cos(2017x).\cos x dx - \int_{0}^{a}\sin^{2017}x. \sin (2017x)dx.$

Xét $\Large \int_{0}^{a} \sin^{2016}x.\cos(2017x).\cos x dx.$

Đặt $\Large \begin{cases}u = \cos(2017x) \\ du = \sin^{2016}x.\cos x dx\end{cases}$

$\Large \Rightarrow \begin{cases}du = -2017 \sin(2017x)dx \\ v = \dfrac{1}{2017} \sin^{2017}x \end{cases}$

Khi đó:

$\Large J = \cos(2017x).\dfrac{1}{2017}. \sin^{2017}x \bigg|_{0}^{a}$ + $\Large \int_{0}^{a}\sin^{2017}x. \sin(2017x)dx.$

Suy ra:

$\Large I = \cos(2017x).\dfrac{1}{2017}. \sin^{2017}x \bigg|_{0}^{a}$ + $\Large \int_{0}^{a}\sin^{2017}x. \sin(2017x)dx$ - $\Large \int_{0}^{a}\sin^{2017}x. \sin(2017x)dx$

$\Large = \cos(2017x).\dfrac{1}{2017}. \sin^{2017}x \bigg|_{0}^{a}$

$\Large = \dfrac{1}{2017}. \sin^{2017}a. \cos(2017a).$