Tính tích phân $I = \int\limits_0^{\dfrac{\pi }{2}} {{{\left( {1 - \cos x} \right)}^n}\sin xdx}$ bằng:

$\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 0$      

$\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 1$

$\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  =  - 1$  

$\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 2$

$\int\limits_{ - 1}^2 {f\left( {2x} \right)} d{\rm{x  =  2}}$

$\int\limits_{ - 3}^3 {f\left( {x + 1} \right)} d{\rm{x  =  2}}$

$\int\limits_{ - 1}^2 {f\left( {2x} \right)} d{\rm{x  =  1}}$

$\int\limits_0^6 {\dfrac{1}{2}f\left( {x - 2} \right)} d{\rm{x  =  1}}$

$\int\limits_{ - a}^a {f\left( x \right)dx}  = 0$

$\int\limits_{ - a}^a {f\left( x \right)dx}  = 1$

$\int\limits_{ - a}^a {f\left( x \right)dx}  =  - 1$

$\int\limits_{ - a}^a {f\left( x \right)dx}  = a$

$I = \dfrac{{ - 1}}{2}$

$I =  - \dfrac{1}{4}$

 $I=\dfrac{1}{4}$

$I =  - 2$

$I =  - \dfrac{1}{4}{\pi ^4}$

$I =  - {\pi ^4}$         

$I = 0$

$I =  - \dfrac{1}{4}$

$I = 2\int\limits_8^9 {\sqrt u du}$

$I = \dfrac{1}{2}\int\limits_8^9 {\sqrt u du}$       

$I = \int\limits_9^8 {\sqrt u du}$

$I = \int\limits_8^9 {\sqrt u du}$

$I = \left. {\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2$                  

$I = \left. {\dfrac{4}{3}\left( {{u^3} + u} \right)} \right|_1^2$

$I = \left. {2\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2$

$I = \left. {\dfrac{1}{3}\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2$