Sử dụng phương pháp đổi biến số để tính tích phân

Đặt $u = \sqrt {3\tan x + 1}  \Rightarrow {u^2} = 3\tan x + 1 \Leftrightarrow 2udu = \dfrac{3}{{{{\cos }^2}x}}dx \Rightarrow \dfrac{{dx}}{{{{\cos }^2}x}} = \dfrac{{2udu}}{3}$

Và $\tan x = \dfrac{{{u^2} - 1}}{3}$

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow u = 1\\x = \dfrac{\pi }{4} \Rightarrow u = 2\end{array} \right.$

Khi đó ta có: $I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{6\tan x}}{{{{\cos }^2}x\sqrt {3\tan x + 1} }}dx}  \Rightarrow I = \int\limits_1^2 {\dfrac{{2\left( {{u^2} - 1} \right)2udu}}{{3u}}}  = \dfrac{4}{3}\int\limits_1^2 {\left( {{u^2} - 1} \right)du}$

Đặt $t = 1 - \cos x \Rightarrow dt = \sin xdx$

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \dfrac{\pi }{2} \Rightarrow t = 1\end{array} \right.$

Khi đó $I = \int\limits_0^1 {{t^n}dt}  = \left. {\dfrac{{{t^{n + 1}}}}{{n + 1}}} \right|_0^1 = \dfrac{1}{{n + 1}}$

Ta có $3\sin x + \cos x = A\left( {2\sin x + 3\cos x} \right) + B\left( {2\cos x - 3\sin x} \right)$

    $\begin{array}{l} \Leftrightarrow 3\sin x + \cos x = \left( {2A - 3B} \right)\sin x + \left( {3A + 2B} \right)\cos x\\ \Leftrightarrow \left\{ \begin{array}{l}2A - 3B = 3\\3A + 2B = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = \dfrac{9}{{13}}\\B =  - \dfrac{7}{{13}}\end{array} \right.\end{array}$

Nên $3\sin x + \cos x = \dfrac{9}{{13}}\left( {2\sin x + 3\cos x} \right) - \dfrac{7}{{13}}\left( {2\cos x - 3\sin x} \right)$

Từ đó ta có

$\begin{array}{l}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{3\sin x + \cos x}}{{2\sin x + 3\cos x}}dx = } \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\dfrac{9}{{13}}\left( {2\sin x + 3\cos x} \right) - \dfrac{7}{{13}}\left( {2\cos x - 3\sin x} \right)}}{{2\sin x + 3\cos x}}dx = } \\ = \dfrac{9}{{13}}\int\limits_0^{\dfrac{\pi }{2}} {dx}  - \dfrac{7}{{13}}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{2\cos x - 3\sin x}}{{2\sin x + 3\cos x}}dx} \\ = \dfrac{{9\pi }}{{26}} - \dfrac{7}{{13}}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{2\sin x + 3\cos x}}d\left( {2\sin x + 3\cos x} \right)} \\ = \dfrac{{9\pi }}{{26}} - \dfrac{7}{{13}}\left. {\ln \left| {2\sin x + 3\cos x} \right|} \right|_0^{\dfrac{\pi }{2}}\\ = \dfrac{{9\pi }}{{26}} - \dfrac{7}{{13}}\ln 2 + \dfrac{7}{{13}}\ln 3\end{array}$

Suy ra $b = \dfrac{7}{{13}};c = \dfrac{9}{{26}} \Rightarrow \dfrac{b}{c} = \dfrac{{14}}{9}$.

Ta có $\int\limits_{ - 1}^1 {f\left( {\left| {2x - 1} \right|} \right)dx}  = \int\limits_{ - 1}^{\frac{1}{2}} {f\left( {1 - 2x} \right)dx}  + \int\limits_{\frac{1}{2}}^1 {f\left( {2x - 1} \right)dx}$

$\Rightarrow I =  - \frac{1}{2}\int\limits_{ - 1}^{\frac{1}{2}} {f\left( {1 - 2x} \right)d\left( {1 - 2x} \right)}  + \frac{1}{2}\int\limits_{\frac{1}{2}}^1 {f\left( {2x - 1} \right)d\left( {2x - 1} \right)}$

$\begin{array}{l} \Leftrightarrow I =  - \frac{1}{2}\int\limits_3^0 {f\left( t \right)dt}  + \frac{1}{2}\int\limits_0^1 {f\left( t \right)dt} \\ \Leftrightarrow I = \frac{1}{2}\int\limits_0^3 {f\left( t \right)dt}  + \frac{1}{2}\int\limits_0^1 {f\left( t \right)dt}  = \frac{1}{2}\left( {2 + 6} \right) = 4\end{array}$

Xét tích phân $\int\limits_3^{2017} {xf\left( x \right)dx}$.

Đặt $x = 2020 - t \Rightarrow dx =  - dt$.

Đổi cận: $\left\{ \begin{array}{l}x = 3 \Rightarrow t = 2017\\x = 2017 \Rightarrow t = 3\end{array} \right.$ , khi đó ta có:

$\begin{array}{l}\int\limits_3^{2017} {xf\left( x \right)dx}  =  - \int\limits_{2017}^3 {\left( {2020 - t} \right)f\left( {2020 - t} \right)dt} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_3^{2017} {\left( {2020 - x} \right)f\left( {2020 - x} \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_3^{2017} {\left( {2020 - x} \right)f\left( x \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2020\int\limits_3^{2017} {f\left( x \right)dx}  - \int\limits_3^{2017} {xf\left( x \right)dx} \\ \Leftrightarrow 2\int\limits_3^{2017} {xf\left( x \right)dx}  = 2020\int\limits_3^{2017} {f\left( x \right)dx} \\ \Leftrightarrow \int\limits_3^{2017} {xf\left( x \right)dx}  = 1010.4\\ \Leftrightarrow \int\limits_3^{2017} {xf\left( x \right)dx}  = 4040\end{array}$

Xét tích phân $\int\limits_1^9 {\dfrac{{f\left( {\sqrt x } \right)}}{{\sqrt x }}{\rm{d}}x = 4}$.

Đặt $t = \sqrt x  \Rightarrow {t^2} = x \Rightarrow 2tdt = dx$.

Đổi cận: $\left\{ \begin{array}{l}x = 1 \Rightarrow t = 1\\x = 9 \Rightarrow t = 3\end{array} \right.$.

Khi đó ta có: $\int\limits_1^9 {\dfrac{{f\left( {\sqrt x } \right)}}{{\sqrt x }}{\rm{d}}x}  = \int\limits_1^3 {\dfrac{{f\left( t \right)2tdt}}{t}}  = 2\int\limits_1^3 {f\left( t \right)dt}  = 2\int\limits_1^3 {f\left( x \right)dx}$.

$\Rightarrow 2\int\limits_1^3 {f\left( x \right)dx}  = 4 \Leftrightarrow \int\limits_1^3 {f\left( x \right)dx}  = 2$.

Xét tích phân $\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)\cos x{\rm{d}}x = 2}$.

Đặt $u = \sin x \Rightarrow du = \cos xdx$.

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow u = 0\\x = \dfrac{\pi }{2} \Rightarrow u = 1\end{array} \right.$.

Khi đó ta có: $\int\limits_0^{\dfrac{\pi }{2}} {f\left( {\sin x} \right)\cos x{\rm{d}}x}  = \int\limits_0^1 {f\left( u \right)du}  = \int\limits_0^1 {f\left( x \right)dx}  = 2$.

Vậy $I = \int\limits_0^3 {f\left( x \right){\rm{d}}x}  = \int\limits_0^1 {f\left( x \right){\rm{d}}x}  + \int\limits_1^3 {f\left( x \right){\rm{d}}x}  = 2 + 2 = 4$.

Đặt $t = \sin 2x \Rightarrow dt = 2\cos 2xdx$ $\Rightarrow  - \dfrac{1}{2}dt = \left( {2{{\sin }^2}x - 1} \right)dx$

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \dfrac{\pi }{4} \Rightarrow t = 1\end{array} \right..$

$\Rightarrow I = \int\limits_0^{\frac{\pi }{4}} {\left( {2{{\sin }^2}x - 1} \right)f\left( {\sin 2x} \right)dx}$$=  - \dfrac{1}{2}\int\limits_0^1 {f\left( t \right)dt}  =  - \dfrac{1}{2}.1 =  - \dfrac{1}{2}.$

Ta có

$\begin{array}{l}f\left( x \right) = \sqrt {x + 2}  + xf\left( {3 - {x^2}} \right)\\ \Rightarrow I = \int\limits_{ - 1}^2 {f\left( x \right)dx}  = \int\limits_{ - 1}^2 {\sqrt {x + 2} dx}  + \int\limits_{ - 1}^2 {xf\left( {3 - {x^2}} \right)dx} \\ \Rightarrow I = {I_1} + {I_2}\end{array}$

Xét tích phân ${I_1} = \int\limits_{ - 1}^2 {\sqrt {x + 2} dx}$.

Đặt $t = \sqrt {x + 2}$ $\Rightarrow {t^2} = x + 2 \Rightarrow 2tdt = dx$.

Đổi cận: $\left\{ \begin{array}{l}x =  - 1 \Rightarrow t = 1\\x = 2 \Rightarrow t = 2\end{array} \right.$.

$\Rightarrow {I_1} = \int\limits_1^2 {t.2tdt}  = 2\int\limits_1^2 {{t^2}dt}  = \left. {\dfrac{{2{t^3}}}{3}} \right|_1^2 = \dfrac{{14}}{3}.$

Xét tích phân ${I_2} = \int\limits_{ - 1}^2 {xf\left( {3 - {x^2}} \right)dx}$.

Đặt $u = 3 - {x^2} \Rightarrow du =  - 2xdx$ $\Rightarrow xdx =  - \dfrac{1}{2}du$.

Đổi cận: $\left\{ \begin{array}{l}x = 1 \Rightarrow u = 2\\x = 2 \Rightarrow u =  - 1\end{array} \right.$.

$\Rightarrow {I_2} = \int\limits_2^{ - 1} { - \dfrac{1}{2}f\left( u \right)du}  = \dfrac{1}{2}\int\limits_{ - 1}^2 {f\left( x \right)dx}  = \dfrac{1}{2}I$.

Vậy $I = \dfrac{{14}}{3} + \dfrac{1}{2}I \Leftrightarrow \dfrac{1}{2}I = \dfrac{{14}}{3} \Leftrightarrow I = \dfrac{{28}}{3}$.

Ta có: $x.f\left( {{x^3}} \right) + f\left( {{x^2} - 1} \right) = {e^{{x^2}}}$ $\Leftrightarrow {x^2}.f\left( {{x^3}} \right) + xf\left( {{x^2} - 1} \right) = x{e^{{x^2}}}$.

Lấy tích phân tư -1 đến 0 hai vế phương trình ta có:

$\int\limits_{ - 1}^0 {{x^2}.f\left( {{x^3}} \right)dx}  + \int\limits_{ - 1}^0 {xf\left( {{x^2} - 1} \right)dx}  = \int\limits_{ - 1}^0 {x{e^{{x^2}}}dx} \,\,\left( * \right)$.

Xét ${I_1} = \int\limits_{ - 1}^0 {{x^2}.f\left( {{x^3}} \right)dx}$.

Đặt $t = {x^3} \Rightarrow dt = 3{x^2}dx \Rightarrow {x^2}dx = \dfrac{{dt}}{3}$.

Đổi cận: $\left\{ \begin{array}{l}x =  - 1 \Rightarrow t =  - 1\\x = 0 \Rightarrow t = 0\end{array} \right.$, khi đó ta có: ${I_1} = \dfrac{1}{3}\int\limits_{ - 1}^0 {f\left( t \right)dt}  = \dfrac{1}{3}\int\limits_{ - 1}^0 {f\left( x \right)dx}$.

Xét ${I_2} = \int\limits_{ - 1}^0 {xf\left( {{x^2} - 1} \right)dx}$.

Đặt $u = {x^2} - 1 \Rightarrow du = 2xdx \Rightarrow xdx = \dfrac{1}{2}du$.

Đổi cận: $\left\{ \begin{array}{l}x =  - 1 \Rightarrow u = 0\\x = 0 \Rightarrow u =  - 1\end{array} \right.$, khi đó ta có ${I_2} = \dfrac{1}{2}\int\limits_0^{ - 1} {f\left( u \right)du}  =  - \dfrac{1}{2}\int\limits_{ - 1}^0 {f\left( x \right)dx}$.

Xét ${I_3} = \int\limits_{ - 1}^0 {x{e^{{x^2}}}dx}$

Đặt $v = {x^2} \Rightarrow dv = 2xdx \Rightarrow xdx = \dfrac{1}{2}dv$.

Đổi cận: $\left\{ \begin{array}{l}x =  - 1 \Rightarrow v = 1\\x = 0 \Rightarrow v = 0\end{array} \right.$, khi đó ta có ${I_3} = \dfrac{1}{2}\int\limits_1^0 {{e^v}dv}  = \dfrac{1}{2}\left. {{e^v}} \right|_1^0 = \dfrac{1}{2} - \dfrac{e}{2} = \dfrac{{1 - e}}{2}$.

Thay tất cả vào (*) ta có:

$\begin{array}{l}\dfrac{1}{3}\int\limits_{ - 1}^0 {f\left( x \right)dx}  - \dfrac{1}{2}\int\limits_{ - 1}^0 {f\left( x \right)dx}  = \dfrac{{1 - e}}{2}\\ \Leftrightarrow  - \dfrac{1}{6}\int\limits_{ - 1}^0 {f\left( x \right)dx}  = \dfrac{{1 - e}}{2}\\ \Leftrightarrow \int\limits_{ - 1}^0 {f\left( x \right)dx}  = 3\left( {e - 1} \right)\end{array}$

Ta có: $I = \int\limits_0^\pi  {xf\left( {\sin x} \right)dx}  = \int\limits_0^{\frac{\pi }{2}} {xf\left( {\sin x} \right)dx}  + \int\limits_{\frac{\pi }{2}}^\pi  {xf\left( {\sin x} \right)dx}$

Xét ${I_1} = \int\limits_{\frac{\pi }{2}}^\pi  {xf\left( {\sin x} \right)dx}$, đặt $t = \pi  - x \Rightarrow dt =  - dx$.

Đổi cận: $\left\{ \begin{array}{l}x = \dfrac{\pi }{2} \Rightarrow t = \dfrac{\pi }{2}\\x = \pi  \Rightarrow t = 0\end{array} \right.$.

Khi đó ta có:

$\begin{array}{l}{I_1} =  - \int\limits_{\frac{\pi }{2}}^0 {\left( {\pi  - t} \right)f\left( {\sin \left( {\pi  - t} \right)} \right)\,} dt\\\,\,\,\,\,\, = \int\limits_0^{\frac{\pi }{2}} {\left( {\pi  - t} \right)f\left( {\sin t} \right)\,} dt\\\,\,\,\,\,\, = \int\limits_0^{\frac{\pi }{2}} {\left( {\pi  - x} \right)f\left( {\sin x} \right)\,} dx\\\,\,\,\,\,\, = \pi \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin x} \right)\,} dx - \int\limits_0^{\frac{\pi }{2}} {xf\left( {\sin x} \right)\,} dx\end{array}$     

$\begin{array}{l} \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {xf\left( {\sin x} \right)dx}  + \pi \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin x} \right)\,} dx - \int\limits_0^{\frac{\pi }{2}} {xf\left( {\sin x} \right)\,} dx\\ \Rightarrow I = \pi \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin x} \right)\,} dx = 5\pi .\end{array}$.

Ta có $\int\limits_{0}^{1}{\frac{\pi {{x}^{3}}+{{2}^{x}}+\text{e}{{x}^{3}}{{.2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\int\limits_{0}^{1}{\left( {{x}^{3}}+\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}} \right)\text{d}x}$ $=\left. \frac{{{x}^{4}}}{4} \right|_{0}^{1}+\int\limits_{0}^{1}{\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{4}+\int\limits_{0}^{1}{\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{4}+J.$

Tính $J=\int\limits_{0}^{1}{\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}$.

Đặt $\pi +\text{e}{{.2}^{x}}=t\Rightarrow \text{e}{{.2}^{x}}\ln 2\text{d}x=\text{d}t\Leftrightarrow {{2}^{x}}\text{d}x=\frac{1}{\text{e}.\ln 2}\text{d}t$.

Đổi cận: Khi $x=0$ thì $t=\pi +\text{e}$; khi $x=1$ thì $t=\pi +2\text{e}$.

Khi đó $J=\int\limits_{0}^{1}{\frac{{{2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{\text{e}\ln 2}\int\limits_{\pi +\text{e}}^{\pi +2\text{e}}{\frac{1}{t}\text{d}t}=\frac{1}{\text{e}\ln 2}\left. \ln \left| t \right| \right|_{\pi +\text{e}}^{\pi +2\text{e}}=\frac{1}{\text{e}\ln 2}\ln \left( 1+\frac{\text{e}}{\text{e}+\pi } \right)$.

Suy ra $\int\limits_{0}^{1}{\frac{\pi {{x}^{3}}+{{2}^{x}}+\text{e}{{x}^{3}}{{.2}^{x}}}{\pi +\text{e}{{.2}^{x}}}\text{d}x}=\frac{1}{4}+\frac{1}{\text{e}\ln 2}\ln \left( 1+\frac{\text{e}}{\text{e}+\pi } \right)$$\Rightarrow m=4$, $n=2$, $p=1$.

Vậy $S=7$.

Đặt

$\begin{array}{l}t = \dfrac{{\sqrt {{x^2} + 1} }}{x} \Leftrightarrow {t^2} = \dfrac{{{x^2} + 1}}{{{x^2}}} = 1 + \dfrac{1}{{{x^2}}}\\ \Rightarrow 2tdt =  - \dfrac{2}{{{x^3}}}dx \Rightarrow tdt =  - \dfrac{{dx}}{{{x^3}}}\end{array}$

Và ${t^2}{x^2} = {x^2} + 1 \Rightarrow {x^2}\left( {{t^2} - 1} \right) = 1 \Leftrightarrow {x^2} = \dfrac{1}{{{t^2} - 1}} \Rightarrow \dfrac{{dx}}{x} =  - \dfrac{t}{{{t^2} - 1}}dt$

Đổi cận: $\left\{ \begin{array}{l}x = 1 \Rightarrow t = \sqrt 2 \\x = \sqrt 3  \Rightarrow t = \dfrac{2}{{\sqrt 3 }}\end{array} \right.$

Khi đó ta có: $I =  - \int\limits_{\sqrt 2 }^{\dfrac{2}{{\sqrt 3 }}} {\dfrac{{{t^2}}}{{{t^2} - 1}}dt}$

Đặt $x = \sqrt k \sin t$ $\Rightarrow dx = \sqrt k \cos tdt$.

Đổi cận: $\left\{ \begin{array}{l}x =  - \sqrt k  \Leftrightarrow \sin t =  - 1 \Leftrightarrow t =  - \dfrac{\pi }{2}\\x = \sqrt k  \Leftrightarrow \sin t = 1 \Leftrightarrow t = \dfrac{\pi }{2}\end{array} \right.$.

Khi đó ta có

$\begin{array}{l}{I_k} = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\sqrt {k - k{{\sin }^2}t} .\sqrt k \cos tdt} \\{I_k} = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {k{{\cos }^2}tdt} \\{I_k} = k\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{1 + \cos 2t}}{2}dt} \\{I_k} = \dfrac{k}{2}\left. {\left( {t + \dfrac{1}{2}\sin 2t} \right)} \right|_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}}\\{I_k} = \dfrac{k}{2}\left( {\dfrac{\pi }{2} + \dfrac{1}{2}\sin \pi  + \dfrac{\pi }{2} - \dfrac{1}{2}\sin \left( { - \pi } \right)} \right)\\{I_k} = \dfrac{k}{2}.\pi  = \dfrac{{k\pi }}{2}\end{array}$

$\begin{array}{l} \Rightarrow {I_1} + {I_2} + {I_3} + ... + {I_{12}}\\ = \dfrac{\pi }{2}\left( {1 + 2 + 3 + ... + 12} \right)\\ = \dfrac{\pi }{2}.\dfrac{{12.13}}{2} = 39\pi \end{array}$