Biết \(\int\limits_0^{\frac{\pi }{4}} {\frac{{\sin 4x}}{{\sqrt {{{\cos }^2}x + 1}  + \sqrt {{{\sin }^2}x + 1} }}{\rm{d}}x}  = \frac{{a\sqr

* Hướng dẫn giải

Ta có \(I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin 4x}}{{\sqrt {{{\cos }^2}x + 1}  + \sqrt {{{\sin }^2}x + 1} }}{\rm{d}}x}  = \sqrt 2 \int\limits_0^{\frac{\pi }{4}} {\frac{{2\sin 2x\cos 2x}}{{\sqrt {3 + \cos 2x}  + \sqrt {3 - \cos 2x} }}{\rm{d}}x} .\)

Đặt \(t = \cos 2x \to {\rm{d}}t =  - 2\sin 2x{\rm{d}}x.\) Đổi cận: \(\left\{ \begin{array}{l}
x = 0 \to t = 1\\
x = \frac{\pi }{4} \to t = 0
\end{array} \right..\)

Khi đó \(I =  - \sqrt 2 \int\limits_1^0 {\frac{t}{{\sqrt {3 + t}  + \sqrt {3 - t} }}{\rm{d}}t}  = \sqrt 2 \int\limits_0^1 {\frac{t}{{\sqrt {3 + t}  + \sqrt {3 - t} }}{\rm{d}}t}  = \frac{1}{{\sqrt 2 }}\int\limits_0^1 {\left( {\sqrt {3 + t}  - \sqrt {3 - t} } \right){\rm{d}}t} \)

\( = \frac{1}{{\sqrt 2 }}\left[ {\frac{2}{3}\sqrt {{{\left( {3 + t} \right)}^3}}  + \frac{2}{3}\sqrt {{{\left( {3 - t} \right)}^3}} } \right]\left| {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right. = \frac{{16\sqrt 2  - 12\sqrt 6  + 8}}{6} \to \left\{ \begin{array}{l}
a = 16\\
b =  - 12\\
c = 8
\end{array} \right. \to P = 36.\)