Tính tích phân sau: ∫5cos x−4sin x/(sinx+cosx)^3dx

Question

Tính tích phân sau:

\(\int _{0}^{\frac{\pi }{2} }\frac{5\cos x-4\sin x}{(\sin x+\cos x)^{3} } dx\)

Answer 1

\(\int _{0}^{\frac{\pi }{2} }\frac{5\cos x-4\sin x}{(\sin x+\cos x)^{3} } dx \)

 Gọi \(I=\int _{0}^{\frac{\pi }{2} }\frac{5\cos x-4\sin x}{(\sin x+\cos x)^{3} } dx  \)

 Đặt \(t=\frac{\pi }{2} -x.\) Ta có:

\(I=\int _{0}^{\frac{\pi }{2} }\frac{5\cos (\frac{\pi }{2} -t)-4\sin (\frac{\pi }{2} -t)}{\left(\sin (\frac{\pi }{2} -t)+\cos (\frac{\pi }{2} -t)\right)^{3} } dt= \int _{0}^{\frac{\pi }{2} }\frac{5\sin t-4\cos t}{(\cos t+\sin t)^{3} } dt\) 

 Gọi J\(=\int _{0}^{\frac{\pi }{2} }\frac{5\sin t-4\cos t}{(\sin t+\cos t)^{3} } dt =\int _{0}^{\frac{\pi }{2} }\frac{5\sin x-4\cos x}{(\sin x+\cos x)^{3} } dx \Rightarrow J=I.\)

 Khi đó: \(I+J=2I=\int _{0}^{\frac{\pi }{2} }\frac{5\cos x-4\sin x}{(\sin x+\cos x)^{3} } dx +\int _{0}^{\frac{\pi }{2} }\frac{5\sin x-4\cos x}{(\sin x+\cos x)^{3} } dx \)

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

Suy ra \(2I=1\Leftrightarrow I=\frac{1}{2} .\)