Bài tập 6 trang 127 SGK Giải tích 12

Bài tập 6 trang 127 SGK Giải tích 12

Câu a:

Biến đổi biểu thức dưới dấu tích phân ta có:

\(cos2x.sin^2x=cos2x\frac{1-cos2x}{2}= \frac{1}{2}cos2x -\frac{1}{2}cos^22x\)

\(=\frac{1}{2}cos2x-\frac{1}{4}(1+cos4x)=\frac{1}{2}cos2x - \frac{1}{4}cos4x-\frac{1}{4}\)

Do đó:

\(\int_{0}^{\frac{\pi }{2}}cos2x sin^2x dx=\int_{0}^{\frac{\pi }{2}} \left ( \frac{1}{2}cos2x -\frac{1}{4}cos4x- \frac{1}{4} \right )dx\)

\(=\frac{1}{4}sin2x \bigg|_{0}^{\frac{\pi }{2}}-\frac{1}{16}sin4x \bigg|_{0}^{\frac{\pi }{2}}- \frac{1}{4}x\bigg|_{0}^{\frac{\pi }{2}}=-\frac{\pi }{8}\)

Câu b:

Ta có: \(\left\{\begin{matrix} 2^x-2^{-x}\geq 0 \ \ \forall x\in [0;1]\\ 2^x-2^{-x}\leq 0 \ \ \forall x\in [-1;0) \end{matrix}\right.\)

Do đó: \(\int_{-1}^{1}\left | 2^x-2^{-x} \right |dx= \int_{-1}^{0}(2^{-x}-2^x)dx+ \int_{0}^{1}(2^{x}-2^{-x})dx\)

\(=\left ( -\frac{2^{-x}}{ln2}-\frac{2^x}{ln2} \right ) \bigg|^0_{-1}+ \left ( \frac{2^x}{ln2}+\frac{2^{-x}}{ln2} \right )\bigg|^1_{0}\)

\(=-\frac{1}{ln2}-\frac{1}{ln2}+\frac{2}{ln2}+\frac{1}{2ln2}+ \frac{2}{ln2}+\frac{1}{2ln2}-\frac{1}{ln 2}-\frac{1}{ln2}=\frac{1}{ln2}\)

Câu c:

Ta có:

\(\frac{(x+1)(x+2)(x+3)}{x^2}=x+6x+\frac{6}{x^2}+\frac{11}{x}\)

Do đó: \(\int_{1}^{2}\frac{(x+1)(x+2)(x+3)}{x^2}dx= \int_{1}^{2}\left ( x+6x+\frac{6}{x^2}+\frac{11}{x} \right )dx\)

\(=\left ( \frac{x^2}{2}+6x-\frac{6}{x} +11lnx \right )\Bigg|^2_1\)

\(=2+12-3+11ln2-\frac{1}{2}+6-6=\frac{21}{2}+11ln2\)
Câu d:

Phân tích biểu thức dưới dấu tích phân ta được:

\(\frac{1}{x^2-2x-3}=\frac{1}{(x+1)(x-3)}=\frac{1}{4} \left ( \frac{1}{x-3}-\frac{1}{x+1} \right )\)

Do đó:

\(\int_{0}^{2}\frac{1}{x^2-2x-3}dx=\frac{1}{4}\int_{0}^{2} \left (\frac{1}{x-3}-\frac{1}{x+1} \right )dx\)

\(=\frac{1}{4}ln\left | \frac{x-3}{x+1} \right | \Bigg |^2_0= \frac{1}{4}\left ( ln\frac{1}{3} -ln3 \right )=-\frac{1}{2}ln3\)

Câu e:
Ta có:

\((sinx+cosx)^2=1+sin2x\)

Do đó: \(\int_{0}^{\frac{\pi }{2}}(sinx+cosx)^2dx=\int_{0}^{\frac{\pi }{2}}(1+sin2x)dx\)

\(=\left ( x-\frac{1}{2}cos2x \right )\Bigg |_0^{\frac{\pi }{2}}= \frac{\pi }{2}+\frac{1}{2}+\frac{1}{2}=\frac{\pi }{2}+1\)

Câu g:

\(\int_{0}^{\pi }(x+sinx)^2dx= \int_{0}^{\pi }(x^2+2xsinx+sin^2x)dx\)

\(=\int_{0}^{\pi }(x^2+2xsin x+\frac{1}{2}-\frac{1}{2}cos2x)dx\)

\(=\left ( \frac{x^3}{3} +\frac{1}{2}x -\frac{1}{4}sin2x \right ) \Bigg|^{\pi}_0\)

\(=\frac{\pi ^3}{3}+\frac{\pi }{2}+2\int_{0}^{\pi }x sinx dx\)

Xét \(\int_{0}^{\pi }x sinx dx.\) Áp dụng phương pháp tính tích phân từng phần bằng cách đặt:

\(\left\{\begin{matrix} u=x\\ dv=sinx dx \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=-cosx \end{matrix}\right.\)

Ta được:

\(\int_{0}^{\pi }x sin dx = - x cosx \bigg|^{\pi}_0+ \int_{0}^{\pi }cosx dx= \pi+sinx \bigg|^{\pi}_0=\pi\)

Vậy \(\int_{0}^{\pi } (x+sinx)^2dx=\frac{\pi ^3}{3}+\frac{\pi }{2}+2\pi= \frac{\pi ^3}{3}+\frac{5\pi }{2}\)

-- Mod Toán 12