Đáp án:
Giải thích các bước giải:
$i)\quad I =\displaystyle\int\limits_0^4\dfrac{xdx}{\sqrt{1+2x}}$
Đặt $u =1 + 2x$
$\to du = 2dx$
Đổi cận:
$x\quad \Big|\quad 0\qquad 4$
$\overline{u\quad \Big|\quad 1\qquad 9}$
Ta được:
$\quad I =\dfrac12\displaystyle\int\limits_1^9\dfrac{u -1}{2\sqrt u}du$
$\to I = \dfrac12\displaystyle\int\limits_1^9\left(\dfrac{\sqrt u}{2} - \dfrac{1}{2\sqrt u}\right)du$
$\to I = \dfrac14\displaystyle\int\limits_1^9\sqrt udu - \dfrac14\displaystyle\int\limits_1^9\dfrac{1}{\sqrt u}du$
$\to I = \dfrac{\sqrt{u^3}}{6}\Bigg|_1^9 - \dfrac{\sqrt u}{2}\Bigg|_1^9$
$\to I = \dfrac{10}{3}$
$l)\quad I = \displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\dfrac{dx}{\sin^4x}$
$\to I =-\dfrac{\cot x}{3\sin^2x}\Bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}+ \dfrac23\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\dfrac{1}{\sin^2x}dx$
$\to I = \dfrac23 -\dfrac23\cot x\Bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}$
$\to I = \dfrac43$
$p)\quad I = \displaystyle\int\limits_0^3\dfrac{1}{x^2 +9}dx$
$\to I = \dfrac19\displaystyle\int\limits_0^3\dfrac{1}{\dfrac{x^2}{9} +1}dx$
$\to I = \dfrac13\displaystyle\int\limits_0^3\dfrac{1}{\dfrac{x^2}{9}+1}d\left(\dfrac x3\right)$
$\to I = \dfrac13\arctan\dfrac x3\Bigg|_0^3$
$\to I = \dfrac{\pi}{12}$
$s)\quad I = \displaystyle\int\limits_{\tfrac{\sqrt2}{2}}^1\dfrac{\sqrt{1 - x^2}}{x^2}dx$
Đặt $x =\sin u$
$\to dx = \cos udu$
Đổi cận:
$x\quad \Big|\quad \dfrac{\sqrt2}{2}\qquad 1$
$\overline{u\quad\Big|\quad \dfrac{\pi}{4}\qquad \,\,\,\dfrac{\pi}{2}}$
Ta được:
$\quad I =\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\dfrac{\sqrt{1 -\sin^2u}}{\sin^2u}\cdot\cos udu$
$\to I = \displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\dfrac{\cos^2u}{\sin^2u}du$
$\to I = \displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\left(\dfrac{1}{\sin^2u} - 1\right)du$
$\to I = \displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\dfrac{1}{\sin^2u}du - \displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}du$
$\to I = -\cot u\Bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} - x\Bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}$
$\to I = 1 -\dfrac{\pi}{4}$