Tích phân (phương pháp đổi biến)

Đặt \(t = {e^x} \Rightarrow dt = {e^x}dx\)

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

Khi đó

\(\begin{array}{l}I = \int\limits_{{e^{ - 1}}}^{{e^2}} {\dfrac{{dt}}{{t + 2}}}  = \left. {\ln \left| {t + 2} \right|} \right|_{{e^{ - 1}}}^{{e^2}} = \ln \left( {{e^2} + 2} \right) - \ln \left( {{e^{ - 1}} + 2} \right) = \ln \dfrac{{{e^2} + 2}}{{{e^{ - 1}} + 2}}\\ = \ln \dfrac{{{e^2} + 2}}{{\dfrac{1}{e} + 2}} = \ln \dfrac{{2e + {e^3}}}{{2e + 1}} = \ln \dfrac{{ae + {e^3}}}{{ae + b}} \Rightarrow \left\{ \begin{array}{l}ae + {e^3} = 2e + {e^3}\\ae + b = 2e + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2\\b = 1\end{array} \right.\end{array}\)

Đặt \(t = {\sin ^2}x \Rightarrow dt = 2\sin x\cos xdx \Rightarrow \sin x\cos xdx = \dfrac{1}{2}dt\) và \({\cos ^2}x = 1 - {\sin ^2}x = 1 - t\)

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^{\dfrac{\pi }{2}} {{e^{{{\sin }^2}x}}\sin x{{\cos }^3}x} dx = \int\limits_0^{\dfrac{\pi }{2}} {{e^{{{\sin }^2}x}}co{s^2}x\sin x\cos x} dx = \dfrac{1}{2}\int\limits_0^1 {{e^t}\left( {1 - t} \right)dt} $

Đặt  \(t = \sqrt {1 + {x^3}}  \Rightarrow {t^2} = 1 + {x^3} \Rightarrow 2tdt = 3{x^2}dx \Rightarrow {x^2}dx = \dfrac{2}{3}tdt\)

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

Khi đó ta có:

\(\begin{array}{l}I = \int\limits_1^2 {\dfrac{{dx}}{{x\sqrt {1 + {x^3}} }}}  = \int\limits_1^2 {\dfrac{{{x^2}dx}}{{{x^3}\sqrt {1 + {x^3}} }}}  = \dfrac{2}{3}\int\limits_{\sqrt 2 }^3 {\dfrac{{dt}}{{{t^2} - 1}}}  = \left. {\dfrac{2}{3}.\dfrac{1}{2}\ln \left| {\dfrac{{t - 1}}{{t + 1}}} \right|} \right|_{\sqrt 2 }^3 \\ = \dfrac{1}{3}\left( {\ln \dfrac{1}{2} - \ln \left( {3 - 2\sqrt 2 } \right)} \right) =  - \dfrac{1}{3}\ln 2 - \dfrac{1}{3}\ln {\left( {\sqrt 2  - 1} \right)^2} =  - \dfrac{1}{3}\ln 2 - \dfrac{2}{3}\ln \left( {\sqrt 2  - 1} \right) \\ = a\ln 2 + b\ln \left( {\sqrt 2  - 1} \right) + c \Rightarrow a =  - \dfrac{1}{3}\end{array}\)

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

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

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

Bước 1: Tính \(\int_0^{\frac{\pi }{{18}}} f (\sin (3x))d(\sin 3x) = 9\), \(\int_0^{\frac{1}{2}} f (x)dx\)

Ta có: \(\int_0^{\frac{\pi }{{18}}} f (\sin (3x))\cos (3x)dx\)\( = \frac{1}{3}\int_0^{\frac{\pi }{{18}}} f (\sin (3x))d(\sin 3x) = 3\)

\( \Rightarrow \int_0^{\frac{\pi }{{18}}} f (\sin (3x))d(\sin 3x) = 9\)

\( \Rightarrow \int_0^{\frac{\pi }{{18}}} f (\sin (3x))d(\sin 3x)\)\( = \int_0^{\frac{1}{2}} f (x)dx = 9\)

Bước 2: Tính \(\int_{ - 1}^{\frac{1}{2}} f (x) \cdot dx\) và \(I = \int_{ - 1}^0 f (x)d\)

Lại có \(\int_{\frac{1}{2}}^2 f (1 - x)dx = 4\). Đặt \(u = 1 - x \Rightarrow du =  - dx\).

Đổi cận: \(x = \dfrac{1}{2} \Rightarrow t = \dfrac{1}{2};x = 2 \Rightarrow t =  - 1\).

\( \Rightarrow \int_{\frac{1}{2}}^2 f (1 - x)dx\)\( = \int_{\frac{1}{2}}^{ - 1} f (u) \cdot ( - du)\)\( = \int_{ - 1}^{\frac{1}{2}} f (u) \cdot du = \int_{ - 1}^{\frac{1}{2}} f (x) \cdot dx = 4\)

Khi đó, \(I = \int_{ - 1}^0 f (x)dx\)\( = \int_{ - 1}^{\frac{1}{2}} f (x)dx - \int_0^{\frac{1}{2}} f (x)dx\)\( = 4 - 9 =  - 5\).

Đặt \(x = {t^3} + 3t + 1\) \( \Rightarrow dx = \left( {3{t^2} + 3} \right)dt\).

Đổi cận: \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = 0\\x = 5 \Rightarrow t = 1\end{array} \right.\).

Ta có: \(\int\limits_1^5 {f\left( x \right)dx}  \)\(= \int\limits_0^1 {f\left( {{t^3} + 3t + 1} \right)\left( {3{t^2} + 3} \right)dt}\)

$ = \int\limits_0^1 {\left( {t + 3} \right)\left( {3{t^2} + 3} \right)dt}  = \dfrac{{57}}{4}$