Cho hàm số $\Large f(x)$ liên tục trên $\Large R$ thỏa mãn $\Large f(\

Cho hàm số $\Large f(x)$ liên tục trên $\Large R$ thỏa mãn $\Large f(\tan x)={{\cos }^{4}}x,\forall x\in R$ . Tính $\Large I=\int\limits_{0}^{1}{f(x)dx}$

• A.

  $\Large \dfrac{\pi +2}{8}$

• B.

  $\Large 1$

• C.

  $\Large \dfrac{2+\pi }{4}$

• D.

  $\Large \dfrac{\pi }{4}$

Chọn A

Đặt $\Large t=\tan x$ . Ta có $\Large \dfrac{1}{{{\cos }^{2}}x}=1+{{\tan }^{2}}x=1+{{t}^{2}}$ $\Large \Rightarrow {{\cos }^{4}}x=\dfrac{1}{{{(1+{{t}^{2}})}^{2}}}\Rightarrow f(t)=\dfrac{1}{{{(1+{{t}^{2}})}^{2}}}$

$\Large I=\int\limits_{0}^{1}{f(x)dx=\int\limits_{0}^{1}{\dfrac{1}{{{(1+{{x}^{2}})}^{2}}}dx}}$

Đặt $\Large x=\tan u\Rightarrow dx=(1+{{\tan }^{2}}u)du$

Đổi cận $\Large x=0\Rightarrow u=0;x=1\Rightarrow u=\dfrac{\pi }{4}$

$\Large I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{1+{{\tan }^{2}}u}{{{(1+{{\tan }^{2}}u)}^{2}}}du}$ $\Large =\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{1}{{{\left( \dfrac{1}{{{\cos }^{2}}u} \right)}^{2}}}.\dfrac{1}{{{\cos }^{2}}u}}du$ $\Large =\int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}udu}$ $\Large =\left( \dfrac{1}{2}u+\dfrac{1}{4}\sin 2u \right)\left| \begin{align}  & \dfrac{\pi }{4} \\ & 0 \\ \end{align} \right.$ $\Large =\dfrac{2+\pi }{8}$