Nếu f(x) là hàm lẻ thì:

Ta có:\(\int\limits_0^{2008\pi } {\sin (2008x + \sin x)dx} = \int\limits_0^{2008\pi } {\sin (2008x)\cos (\sin x) + \cos (2008x)\sin (\sin x)dx} \)

Đặt f(x) = sin(sin(x)), ta có f(-x) = sin(sin(-x)) = sin(-sin(x)) = -sin(sin(x)) = -f(x). Vậy f(x) là hàm lẻ.

Đặt g(x) = cos(sin(x)), ta có g(-x) = cos(sin(-x)) = cos(-sin(x)) = cos(sin(x)) = g(x). Vậy g(x) là hàm chẵn.

Ta có:\(\int\limits_0^{2008\pi } {\sin (2008x + \sin x)dx} = \int\limits_0^{2008\pi } {\sin (2008x)\cos (\sin x)dx} + \int\limits_0^{2008\pi } {\cos (2008x)\sin (\sin x)dx} \)

Xét tích phân:\(I = \int\limits_0^{2008\pi } {\cos (2008x)\sin (\sin x)dx} \). Đặt t = 2008π - x. Khi đó dt = -dx, x = 0 ⇒ t = 2008π, x = 2008π ⇒ t = 0.

Do đó:\(I = \int\limits_{2008\pi }^0 {\cos (2008(2008\pi - t))\sin (\sin (2008\pi - t))( - dt)} = \int\limits_0^{2008\pi } {\cos (2008(2008\pi - t))\sin (\sin (2008\pi - t))dt} \)

Vì cos(2008(2008π - t)) = cos(2008.2008π - 2008t) = cos(-2008t) = cos(2008t)

và sin(sin(2008π - t)) = sin(sin(-t)) = sin(-sin(t)) = -sin(sin(t))

Nên:\(I = \int\limits_0^{2008\pi } {\cos (2008t)( - \sin (\sin t))dt} = - \int\limits_0^{2008\pi } {\cos (2008t)\sin (\sin t)dt} = - I\)

Suy ra I = 0.

Xét tích phân:\(J = \int\limits_0^{2008\pi } {\sin (2008x)\cos (\sin x)dx} \)

Đặt x = u + π. Khi đó dx = du, x = 0 ⇒ u = -π, x = 2008π ⇒ u = 2007π.

Suy ra:\(J = \int\limits_{ - \pi }^{2007\pi } {\sin (2008(u + \pi ))\cos (\sin (u + \pi ))du} = \int\limits_{ - \pi }^{2007\pi } {\sin (2008u + 2008\pi )\cos ( - \sin u)du} \)

Vì sin(2008u + 2008π) = sin(2008u)cos(2008π) + cos(2008u)sin(2008π) = sin(2008u)

và cos(-sin(u)) = cos(sin(u))

Nên:\(J = \int\limits_{ - \pi }^{2007\pi } {\sin (2008u)\cos (\sin u)du} \)

Ta có:\(\int\limits_{ - \pi }^{2007\pi } {\sin (2008u)\cos (\sin u)du} = \int\limits_{ - \pi }^0 {\sin (2008u)\cos (\sin u)du} + \int\limits_0^{2007\pi } {\sin (2008u)\cos (\sin u)du} \)

Xét:\(\int\limits_{ - \pi }^0 {\sin (2008u)\cos (\sin u)du} \). Đặt t = -u. Khi đó dt = -du, u = -π ⇒ t = π, u = 0 ⇒ t = 0.

Suy ra:\(\int\limits_{ - \pi }^0 {\sin (2008u)\cos (\sin u)du} = \int\limits_{\pi }^0 {\sin ( - 2008t)\cos (\sin ( - t))( - dt)} = \int\limits_{\pi }^0 {\sin ( - 2008t)\cos ( - \sin (t))( - dt)} \)

Vì sin(-2008t) = -sin(2008t) và cos(-sin(t)) = cos(sin(t))

Nên:\(\int\limits_{\pi }^0 {\sin ( - 2008t)\cos ( - \sin (t))( - dt)} = - \int\limits_0^{\pi } {\sin (2008t)\cos (\sin t)dt} \)

Do đó:\(J = - \int\limits_0^{\pi } {\sin (2008u)\cos (\sin u)du} + \int\limits_0^{2007\pi } {\sin (2008u)\cos (\sin u)du} = \int\limits_0^{2007\pi } {\sin (2008u)\cos (\sin u)du} - \int\limits_0^{\pi } {\sin (2008u)\cos (\sin u)du} \)

Ta có:\(\int\limits_0^{2007\pi } {\sin (2008u)\cos (\sin u)du} = \int\limits_0^{\pi } {\sin (2008u)\cos (\sin u)du} + \int\limits_{\pi }^{2\pi } {\sin (2008u)\cos (\sin u)du} + ... + \int\limits_{2006\pi }^{2007\pi } {\sin (2008u)\cos (\sin u)du} \)

Xét:\(\int\limits_{k\pi }^{(k + 1)\pi } {\sin (2008u)\cos (\sin u)du} \). Đặt u = t + kπ. Khi đó du = dt, u = kπ ⇒ t = 0, u = (k+1)π ⇒ t = π.

Suy ra:\(\int\limits_{k\pi }^{(k + 1)\pi } {\sin (2008u)\cos (\sin u)du} = \int\limits_0^{\pi } {\sin (2008(t + k\pi ))\cos (\sin (t + k\pi ))dt} = \int\limits_0^{\pi } {\sin (2008t + 2008k\pi )\cos (\sin t\cos k\pi + \cos t\sin k\pi )dt} \)

Vì sin(2008t + 2008kπ) = sin(2008t)cos(2008kπ) + cos(2008t)sin(2008kπ) = sin(2008t)cos(2008kπ) = sin(2008t)

Nên:\(\int\limits_{k\pi }^{(k + 1)\pi } {\sin (2008u)\cos (\sin u)du} = \int\limits_0^{\pi } {\sin (2008t)\cos (\sin (t + k\pi ))dt} \)

Do đó:\(J = 0\)

Vậy:\(\int\limits_0^{2008\pi } {\sin (2008x + \sin )dx} = 0\)

Đặt \(t = \sqrt {{e^x} + 1} \Rightarrow {t^2} = {e^x} + 1 \Rightarrow 2tdt = {e^x}dx \Rightarrow dx = \frac{{2tdt}}{{{e^x}}} = \frac{{2tdt}}{{{t^2} - 1}}\)

Đổi cận:

Với \(x = 0 \Rightarrow t = \sqrt 2 \)

Với \(x = \ln 3 \Rightarrow t = 2\)

Khi đó:

\(\int\limits_0^{\ln 3} {\frac{{dx}}{{\sqrt {{e^x} + 1} }}} = \int\limits_{\sqrt 2 }^2 {\frac{{2tdt}}{{\left( {{t^2} - 1} \right)t}}} = 2\int\limits_{\sqrt 2 }^2 {\frac{{dt}}{{{t^2} - 1}}} = 2\int\limits_{\sqrt 2 }^2 {\frac{1}{{2\left( {t - 1} \right)}}} - \frac{1}{{2\left( {t + 1} \right)}}dt\)

\(= \int\limits_{\sqrt 2 }^2 {\frac{1}{{t - 1}}} - \frac{1}{{t + 1}}dt = \ln \left| {t - 1} \right| - \ln \left| {t + 1} \right|} |_{\sqrt 2 }^2 = \ln \left| {\frac{{t - 1}}{{t + 1}}} \right|_{\sqrt 2 }^2 = \ln \frac{1}{3} - \ln \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} = \ln \frac{{\frac{1}{3}}}{{\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}}} = \ln \frac{{\sqrt 2 + 1}}{{3(\sqrt 2 - 1)}}}\)