Đề 003-TN THPT QG

Đề 003-TN THPT QG

Câu 45.  Cho hàm số $f\left( x \right)$ có $f\left( \frac{\pi }{2} \right)=0$ và ${f}’\left( x \right)=\text{cos}2x.{{\sin }^{3}}x$. Khi đó $\int\limits_{0}^{\frac{\pi }{6}}{f\left( x \right)}\text{d}x$ bằng

A. $\frac{253\pi }{1200}$.                                  

B. $\frac{251}{1200}$.           

C. $\frac{251\pi }{1200}$.                        

D. $\frac{253}{1200}$.

Hướng dẫn và Lời giải

Chọn D

Ta có $f\left( x \right)=\int{\text{cos}2x.{{\sin }^{3}}x\text{d}x}=\int{\left( 2\text{co}{{\text{s}}^{2}}x-1 \right)}\left( 1-\text{co}{{\text{s}}^{2}}x \right)\sin x\text{d}x$.

Đặt $t=\text{cos}x\Rightarrow \text{d}t=-\sin x\text{d}x$.

Khi đó: $\int{\left( 2\text{co}{{\text{s}}^{2}}x-1 \right)}\left( 1-\text{co}{{\text{s}}^{2}}x \right)\sin x\text{d}x=-\int{\left( 2{{t}^{2}}-1 \right)\left( 1-{{t}^{2}} \right)\text{d}t}=\int{\left( 2{{t}^{4}}-3{{t}^{2}}+1 \right)\text{d}t}$

$=\frac{2}{5}{{t}^{5}}-{{t}^{3}}+t+C$

$=\frac{2}{5}\text{co}{{\text{s}}^{5}}x-\text{co}{{\text{s}}^{3}}x+\text{cos}x+C$.

Suy ra:$f\left( x \right)=\frac{2}{5}\text{co}{{\text{s}}^{5}}x-\text{co}{{\text{s}}^{3}}x+\text{cos}x+C$

Mà $f\left( \frac{\pi }{2} \right)=0\Leftrightarrow C=0$.

Do đó $f\left( x \right)=\frac{2}{5}\text{co}{{\text{s}}^{5}}x-\text{co}{{\text{s}}^{3}}x+\text{cos}x=\text{cos}x\left( \frac{2}{5}\text{co}{{\text{s}}^{4}}x-\text{co}{{\text{s}}^{2}}x+1 \right)=\text{cos}x\left[ \frac{2}{5}{{\left( 1-\text{si}{{\text{n}}^{2}}x \right)}^{2}}+\text{si}{{\text{n}}^{2}}x \right]$

$I=\int\limits_{0}^{\frac{\pi }{6}}{f\left( x \right)}\text{ d}x=\int\limits_{0}^{\frac{\pi }{6}}{\text{cos}x\left[ \frac{2}{5}{{\left( 1-\text{si}{{\text{n}}^{2}}x \right)}^{2}}+\text{si}{{\text{n}}^{2}}x \right]\text{d}x}$.

Đặt $t=\sin x\Rightarrow \text{d}t=\text{cos}x\text{d}x$.

Đổi cận

$\left\{ \begin{array}{l} x = 0 \Rightarrow t = 0\\ x = \frac{\pi }{6} \Rightarrow t = \frac{1}{2} \end{array} \right.$

$I=\int\limits_{0}^{\frac{1}{2}}{\left[ \frac{2}{5}{{\left( 1-{{t}^{2}} \right)}^{2}}+{{t}^{2}} \right]}\text{d}t=\frac{1}{5}\int\limits_{0}^{\frac{1}{2}}{\left( 2{{t}^{4}}+{{t}^{2}}+2 \right)\text{d}t}=\frac{1}{5}\left. \left( \frac{2}{5}{{t}^{5}}+\frac{1}{3}{{t}^{3}}+2t \right) \right|_{0}^{\frac{1}{2}}$$=\frac{253}{1200}$.

Trở lại đề thi