em cung da tung thay tinh chac nay$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$Sử dụng tính chất $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$,ta có:
$$I_{13}=\int\limits_{0}^{\frac{\pi}{2}} {\ln \left[\frac{ \left(1+\cos{x} \right)^{1+\cos{x}}}{1+\sin{x}} \right]dx}=\int\limits_{0}^{\frac{\pi}{2}} \ln \left[\frac{(1+\sin{x})^{1+\sin{x}}}{1+\cos{x}} \right]dx$$
Suy ra:
$$2I_{13}=\int\limits_{0}^{\frac{\pi}{2}} \ln \left[(1+\cos{x})^{\cos{x}}(1+\sin{x})^{\sin{x}} \right]dx=\int\limits_{0}^{\frac{\pi}{2}} \cos{x}\ln{(1+\cos{x})}dx+\int\limits_{0}^{\frac{\pi}{2}}\sin{x}\ln{(1+\sin{x})}dx$$
Đặt $J=\int\limits_{0}^{\frac{\pi}{2}} \cos{x}\ln{(1+\cos{x})}dx;K=\int\limits_{0}^{\frac{\pi}{2}}\sin{x}\ln{(1+\sin{x})}dx$.
*Với J,sử dụng công thức tích phân từng phần,ta có:
$$J=\int\limits_{0}^{\frac{\pi}{2}}\ln{(1+\cos{x})}d(\sin{x})=\left[\sin{x}.\ln{(1+\cos{x})} \right]\Big|_{0}^{\frac{\pi}{2}}+\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2{x}}{1+\cos{x}}dx$$
Xét :
$$J_1=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2{x}}{1+\cos{x}}dx=2\int\limits_{0}^{\frac{\pi}{2}}\sin^2{\frac{x}{2}}dx=x\Big|_{0}^{\frac{\pi}{2}}-\sin{x}\Big|_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}-1$$
Suy ra:
$$J=\frac{\pi}{2}-1$$
*Với K,cũng sử dụng công thức tích phân từng phần,ta có:
$$K=\int\limits_{0}^{\frac{\pi}{2}}\ln{(1+\sin{x})}d(-\cos{x})=\left[-\cos{x}.\ln{(1+\sin{x})} \right]\Big|_{0}^{\frac{\pi}{2}}+\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^2{x}}{1+\sin{x}}dx=J_1=\frac{\pi}{2}-1$$
Vậy:
$$2I_{13}=J+K=2\left(\frac{\pi}{2}-1 \right) \Rightarrow I_{13}=\frac{\pi}{2}-1$$
P/s:@Anh Thành:Anh rảnh thì post lời giải câu 10 giùm em,cảm ơn
nhung ko hieu cach ap dung