Question

Tính các tích phân sau:

a) \(\int_0^1x^3\sqrt{1-x^2}dx\)

b) \(\int_1^2\dfrac{dx}{x^2-2x+2}\)

c) \(\int_1^2\dfrac{dx}{\sqrt{4-x^2}}\)

d) \(\int_0^1x\sqrt{x^2+1}dx\)

Answer 1

a.

Đặt \(\sqrt{1-x^2}=u\Rightarrow x^2=1-u^2\Rightarrow xdx=-udu\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^0_1\left(1-u^2\right).u.\left(-udu\right)=\int\limits^1_0\left(u^2-u^4\right)du=\left(\dfrac{1}{3}u^3-\dfrac{1}{5}u^5\right)|^1_0\)

\(=\dfrac{2}{15}\)

 

Answer 2

b.

\(\int\limits^2_1\dfrac{dx}{x^2-2x+2}=\int\limits^2_1\dfrac{dx}{\left(x-1\right)^2+1}\)

Đặt \(x-1=tanu\Rightarrow dx=\dfrac{1}{cos^2u}du\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=0\\x=2\Rightarrow u=\dfrac{\pi}{4}\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{1}{tan^2u+1}.\dfrac{1}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{cos^2u}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0du\)

\(=u|^{\dfrac{\pi}{4}}_0=\dfrac{\pi}{4}\)

Answer 3

c.

\(\int\limits^2_1\dfrac{dx}{\sqrt{4-x^2}}\)

Đặt \(x=2sinu\Rightarrow dx=2cosu.du\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=\dfrac{\pi}{6}\\x=2\Rightarrow u=\dfrac{\pi}{2}\end{matrix}\right.\)

\(I=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{2cosu.du}{\sqrt{4-4sin^2u}}=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{2cosu.du}{2cosu}=\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}du\)

\(=u|^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}=\dfrac{\pi}{3}\)

Answer 4

d.

\(\int\limits^1_0x\sqrt{x^2+1}dx\)

Đặt \(\sqrt{x^2+1}=u\Rightarrow x^2=u^2-1\Rightarrow xdx=u.du\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=\sqrt{2}\end{matrix}\right.\)

\(I=\int\limits^{\sqrt{2}}_1u.\left(u.du\right)=\int\limits^{\sqrt{2}}_1u^2du=\dfrac{1}{3}u^3|^{\sqrt{2}}_1=\dfrac{2\sqrt{2}-1}{3}\)