Tính tích phân sau: ∫sinx/(sinx+cosx)^3dx

Question

Tính tích phân sau:

\(\int _{0}^{\frac{\pi }{2} }\frac{\sin x}{(\sin x+\cos x)^{3} } dx\)

Answer 1

\(\int _{0}^{\frac{\pi }{2} }\frac{\sin x}{(\sin x+\cos x)^{3} } dx \)

 Gọi \(A=\int _{0}^{\frac{\pi }{2} }\frac{\sin x}{(\sin x+\cos x)^{3} } dx  \)

 Đặt \(t=\frac{\pi }{2} -x. Ta có A=\int _{0}^{\frac{\pi }{2} }\frac{\sin (\frac{\pi }{2} -t)}{\left(\sin (\frac{\pi }{2} -t)+\cos (\frac{\pi }{2} -t)\right)^{3} } dt\)

\(= \int _{0}^{\frac{\pi }{2} }\frac{\cos t}{\left(\cos t+\sin t\right)^{3} } dt \)

 Gọi \(B=\int _{0}^{\frac{\pi }{2} }\frac{\cos x}{\left(\cos x+\sin x\right)^{3} } dt \Rightarrow B=A\)

 Khi đó: \(A+B=2A=\int _{0}^{\frac{\pi }{2} }\frac{\sin x}{(\sin x+\cos x)^{3} } dx +\int _{0}^{\frac{\pi }{2} }\frac{\cos x}{(\cos x+\sin x)^{3} } dx\)

\(=\int _{0}^{\frac{\pi }{2} }\frac{1}{\left(\cos x+\sin x\right)^{2} } dx .\)
\(=\int _{0}^{\frac{\pi }{2} }\frac{1}{2\sin ^{2} (x+\frac{\pi }{4} )} dx =-\frac{1}{2} \left. \cot (x+\frac{\pi }{4} )\right|_{0}^{\frac{\pi }{2} } =1 \)
Suy ra \(2A=1\Leftrightarrow A=\frac{1}{2} .\)