Chơi tổng quát cho nó máu nha
$I=\int \dfrac{1}{\cos^{2n+1}x }dx=\int \dfrac{d(x+\dfrac{\pi}{2})}{\sin^{2n+1} (x+\dfrac{\pi}{2})}=\int \dfrac{1}{\sin^{2n+1} u}du$
$=\dfrac{1}{2^{2n+1}}\int \dfrac{1}{\bigg (\sin \dfrac{u}{2} \cos \dfrac{u}{2} \bigg )^{2n+1}}du=\dfrac{1}{2^{2n+1}}\int \dfrac{1}{\bigg (\tan \dfrac{u}{2} \bigg )^{2n+1} \bigg ( \cos \dfrac{u}{2} \bigg )^{4n+2} }du$
$=\dfrac{1}{2^{2n}}\int \dfrac{ \bigg (1+\tan^2 \dfrac{u}{2} \bigg )^{2n} d(\tan \dfrac{u}{2}) }{ {\bigg (\tan \dfrac{u}{2} \bigg )^{2n+1} }}$
Đến đây khai triển $\bigg (1+\tan^2 \dfrac{u}{2} \bigg )^{2n} $ ra rồi chia đa thức là xong
Để bạn trẻ dễ hiểu thì đặt $\tan \dfrac{u}{2}=t$ ta có
$I=\dfrac{1}{2^{2n}} \int \dfrac{(1+u^2)^{2n}}{u^{2n+1}}du$ XONG NHÁ