cho hàm số f(x) có f(pi/2)=0 và f'(x)=sinx.sin^2(2x). Khi đó tích phân từ 0 đến pi/2 của f(x)dx bằng

Đáp án:

$\displaystyle\int\limits_0^{\pi /2} {f\left( x \right)dx} = \dfrac{{ - 104}}{{225}}$

Lời giải:

$\begin{array}{l}
f'\left( x \right) = \sin x.{\left( {\sin 2x} \right)^2}\\
 = \sin x.{\left( {2\sin x.\cos x} \right)^2}\\
 = 4.{\sin ^3}x.{\cos ^2}x\\
 = 4.{\cos ^2}x.{\sin ^2}x.\sin x\\
 = 4.\cos^2x.\left( {1 - {{\cos }^2}x} \right).\sin x\\
 = \left( {4.{{\cos }^4}x - 4{{\cos }^2}x} \right).\left( { - \sin x} \right)\\
 \Rightarrow f\left( x \right) = \displaystyle\int {\left( {4{{\cos }^4}x - 4{{\cos }^2}x} \right)\left( { - \sin x} \right)dx} \\
 = \displaystyle\int {\left( {4{{\cos }^4}x - 4{{\cos }^2}x} \right)d\cos x} \\
 = \dfrac{4}{5}{\cos ^5}x - \dfrac{4}{3}{\cos ^3}x + C\\
Do:f\left( {\dfrac{\pi }{2}} \right) = 0 \Rightarrow C = 0\\
 \Rightarrow f\left( x \right) = \dfrac{4}{5}{\cos ^5}x - \dfrac{4}{3}{\cos ^3}x\\
 \Rightarrow \displaystyle\int\limits_0^{\pi /2} {f\left( x \right)dx} \\
 = \displaystyle\int\limits_0^{\pi /2} {\left( {\dfrac{4}{5}{{\cos }^5}x - \dfrac{4}{3}{{\cos }^3}x} \right)dx} \\
 = \dfrac{4}{5}\displaystyle\int\limits_0^{\pi /2} {{{\cos }^4}x.\cos xdx}  - \dfrac{4}{3}\displaystyle\int\limits_0^{\pi /2} {{{\cos }^2}x.\cos xdx} \\
 = \dfrac{4}{5}.\displaystyle\int\limits_0^{\pi /2} {{{\left( {1 - {{\sin }^2}x} \right)}^2}d\left( {\sin x} \right)}  - \dfrac{4}{3}.\displaystyle\int\limits_0^{\pi /2} {1 - {{\sin }^2}xd\left( {\sin x} \right)} \\
 = \dfrac{4}{5}.\left( {\sin x - \dfrac{2}{3}{{\sin }^3}x + \dfrac{1}{5}{{\sin }^5}x} \right)\Bigg\vert_0^{\pi /2} - \dfrac{4}{3}.\left( {\sin x - \dfrac{{{{\sin }^3}x}}{3}} \right)\Bigg\vert_0^{\pi /2}\\
 = \dfrac{4}{5}.\dfrac{8}{{15}} - \dfrac{4}{3}.\dfrac{2}{3}\\
 = \dfrac{{ - 104}}{{225}}
\end{array}$