Lời giải:
Ta có:
\(I=\int (\sin x)^{2016}\cos (2018x)dx=\int (\sin x)^{2016}\cos (2017x+x)dx\)
\(=\int \sin ^{2016}x\cos (2017x)\cos xdx-\int \sin ^{2017}x\sin (2017x)dx\)
(Khai triển theo công thức lượng giác \(\cos (a+b)=\cos a\cos b-\sin a\sin b\) )
Thực hiện nguyên hàm từng phần:
\(\left\{\begin{matrix} u=\cos (2017x)\\ dv=\sin ^{2016}x\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=-2017\sin (2017x)dx\\ v=\int \sin ^{2016}x\cos xdx=\int \sin ^{2016}xd(\sin x)=\frac{\sin ^{2017}x}{2017}\end{matrix}\right.\)
\(\Rightarrow \int \sin ^{2016}x\cos (2017x)\cos xdx=\frac{\sin ^{2017}x\cos (2017x)}{2017}+\int \sin ^{2017}x\sin (2017x)dx \)
Suy ra:
\(I=\frac{\sin ^{2017}x\cos (2017x)}{2017}+\int \sin ^{2017}x\cos (2017x)dx-\int \sin ^{2017}x\cos (2017x)dx\)
\(=\frac{\sin ^{2017}x\cos (2017x)}{2017}\)
\(\Rightarrow \int ^{a}_{0}\sin ^{2016}x\cos (2018x)dx=\frac{\sin ^{2017}a\cos (2017a)}{2017}\)
