Bài tập 3 trang 113 SGK Giải tích 12

Câu a:

Đặt u= x+1 ta có du =  dx; x² = (u - 1)²

Khi x = 0 thì u = 1; khi x = 3 thì u = 4. Khi đó :

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

\(\int_{1}^{4}(u^{\frac{1}{2}}-2u^{\frac{1}{2}}+u^{\frac{3}{2}})du= \int_{1}^{4}u^{\frac{1}{2}}du -2\int_{1}^{4}u^{\frac{1}{2}}du+ \int_{1}^{4}u^{\frac{3}{2}}du\)

\(=\frac{2}{3}u^{\frac{3}{2}} \Bigg| ^4_1-4u^{\frac{1}{2}}\Bigg| ^4_1- 2u\Bigg| ^4_1=\frac{16}{3}-\frac{2}{3}-(8-4)-2(\frac{1}{2}-1)\)

\(=\frac{14}{3}-3=\frac{5}{3}\)

Câu b:

Đặt x = sint ta có: dx = costdt

Khi x = 0 thì t = 0; khi x = 1 thì \(t=\frac{\pi }{2}\). Do đó:

\(\int_{0}^{1}\sqrt{1-x^2}dx=\int_{0}^{\frac{\pi }{2}}\sqrt{1-sin^2 t}cos tdt\)

\(=\int_{0}^{\frac{\pi }{3}}\left | cos t \right |.cos t dt = \int_{0}^{\frac{\pi }{3}}cos^2 t dt\) (Vì \(cos t \geq 0, \forall t \in \left [ 0; \frac{\pi }{2} \right ]\))

\(\Rightarrow \int_{0}^{1}\sqrt{1-x^2}dx = \int_{0}^{\frac{\pi }{2}}cos^2t dt= \frac{1}{2} \int_{0}^{\frac{\pi }{2}} (1+cos 2t)dt\)

\(=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}dt +\frac{1}{4}\int_{0}^{\frac{\pi }{2}}cos 2t dt =\frac{1}{2}t \Bigg|_0^{\frac{\pi }{2}}+\frac{1}{4}sin 2t \Bigg|_0^{\frac{\pi }{2}}\)

\(=\frac{\pi }{4}+0= \frac{\pi }{4}\)

Vậy \(\int_{0}^{1}\sqrt{1-x^2}dx=\frac{\pi }{4}.\)

Câu c:

Đặt \(u=x.e^x\) ta có: \(du=(e^x+x.e^x)dx=e^x(x+1)dx\)

Khi x = 0 thì u = 0; khi x = 1 thì u = e

Do đó: \(\int_{0}^{1}\frac{e^x(x+1)}{1+x.e^x}dx=\int_{0}^{e}\frac{du}{1+u}= ln(1+u) \Bigg|^e_0=ln(e+1)\)

Câu d:

Đặt x = a sint ta có: dx = acost dt

Khi x = 0 thì t = 0; khi \(x=\frac{a}{2}\) thì \(t=\frac{\pi }{6}\). Do đó:

\(\int_{0}^{\frac{a}{2}}\frac{1}{\sqrt{a^2-x^2}}dx= \int_{0}^{\frac{\pi }{6}}\frac{acos t dt}{\sqrt{a^2-a^2sin^2t}}\)

\(=\int_{0}^{\frac{\pi }{6}}\frac{a.cost dt}{\left | a cos t \right |}\)

Vì a > 0 và \(cos t \geq 0, \forall t \in \left [ 0; \frac{\pi }{6} \right ]\)

Nên \(\int_{0}^{\frac{\pi }{6}}\frac{a.cos t .dt}{a.cost}= \int_{0}^{\frac{\pi }{6}}dt =t \Bigg|^{\frac{\pi }{6}}_0=\frac{\pi }{6}\)

Vậy \(\int_{0}^{\frac{a}{2}}\frac{1}{\sqrt{a^2-x^2}}dx=\frac{\pi }{6}.\)

-- Mod Toán 12 HỌC247

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