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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;\dfrac{\pi }{2} \right]$ và $f\left( x \right)+f\left( \dfrac{\pi }{2}-x \right)=\dfrac{\cos x}{{{\left( 1+\sin x \right)}^{2}}},\forall x\in \left[ 0;\dfrac{\pi }{2} \right]$. Tính tích phân $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}$.
A. $I=\dfrac{1}{4}$
B. $I=1$
C. $I=\dfrac{1}{2}$
D. $I=2$
A. $I=\dfrac{1}{4}$
B. $I=1$
C. $I=\dfrac{1}{2}$
D. $I=2$
Xét tích phân ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}$. Đặt $u=\dfrac{\pi }{2}-x\Rightarrow du=-dx$.
Đổi cận $x=0\Rightarrow u=\dfrac{\pi }{2};\ x=\dfrac{\pi }{2}\Rightarrow u=0$.
Suy ra $\begin{aligned}
& {{I}_{1}}=-\int\limits_{\dfrac{\pi }{2}}^{0}{f\left( \dfrac{\pi }{2}-x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx}\Rightarrow 2{{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx} \\
& \Rightarrow 2{{I}_{1}}=\left( \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)}+f\left( \dfrac{\pi }{2}-x \right) \right)dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{{{\left( 1+\sin x \right)}^{2}}}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\left( 1+\sin x \right)}{{{\left( 1+\sin x \right)}^{2}}}} \\
& =-\dfrac{1}{1+\sin x}\left| _{0}^{\dfrac{\pi }{2}} \right.=-\left( \dfrac{1}{2}-1 \right)=\dfrac{1}{2}\Rightarrow {{I}_{1}}=\dfrac{1}{4} \\
\end{aligned}$
Đổi cận $x=0\Rightarrow u=\dfrac{\pi }{2};\ x=\dfrac{\pi }{2}\Rightarrow u=0$.
Suy ra $\begin{aligned}
& {{I}_{1}}=-\int\limits_{\dfrac{\pi }{2}}^{0}{f\left( \dfrac{\pi }{2}-x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx}\Rightarrow 2{{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx} \\
& \Rightarrow 2{{I}_{1}}=\left( \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)}+f\left( \dfrac{\pi }{2}-x \right) \right)dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{{{\left( 1+\sin x \right)}^{2}}}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\left( 1+\sin x \right)}{{{\left( 1+\sin x \right)}^{2}}}} \\
& =-\dfrac{1}{1+\sin x}\left| _{0}^{\dfrac{\pi }{2}} \right.=-\left( \dfrac{1}{2}-1 \right)=\dfrac{1}{2}\Rightarrow {{I}_{1}}=\dfrac{1}{4} \\
\end{aligned}$
Đáp án A.