Chủ đề này có 3 trả lời

[vietfrog]

Tìm nguyên hàm:
$\begin{array}{l}a.\int {\dfrac{{dx}}{{\sqrt {{x^2} + k} }}} \\b.\int {\sqrt {{x^2} + k} .dx} \end{array}$
( Chứng minh bằng cách biến đổi nhé, không được đạo hàm đâu đấy! )

Tính :
$\begin{array}{l}a.\int\limits_1^2 {\dfrac{{\sqrt {1 + {x^2}} }}{{{x^4}}}dx} \\b.\int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + c{\rm{o}}{{\rm{s}}^2}x}}dx} \end{array}$

Mình mới tự cầy nguyên hàm tích phân nên không thạo lắm, mọi người giải chi tiết nhé!!!

[stuart clark]

(a)$\; \displaystyle \int_{1}^{2}\dfrac{\sqrt{1+x^2}}{x^4}dx$

$\; =\displaystyle \int_{1}^{2}\dfrac{x.\sqrt{\dfrac{1}{x^2}+1}}{x^4}dx $

$\;=\displaystyle \int_{1}^{2}\dfrac{\sqrt{\dfrac{1}{x^2}+1}}{x^3}dx$

let $\displaystyle \dfrac{1}{x^2}+1=t^2\Leftrightarrow -\dfrac{2}{x^3}dx=2tdt\Leftrightarrow \dfrac{1}{x^3}dx=-tdt$

$=-\displaystyle\int_{1}^{2}t^2dt=-\dfrac{t^3}{3}\Big|_{1}^{2}=-\dfrac{1}{3}.\left(\dfrac{1}{x^2}+1\right)^{\dfrac{3}{2}} \Big|_{1}^{2}=-\dfrac{1}{3}\left\{\left(\dfrac{5}{4}\right)-2\right\}$



(b) Let $\displaystyle I = \int_{0}^{\pi}\dfrac{x \sin x}{1+\cos^2 x}dx$

$\displaystyle I = \int_{0}^{\pi}\dfrac{(\pi-x) \sin (\pi-x)}{1+\cos^2 (\pi-x)}dx=\int_{0}^{\pi}\dfrac{(\pi-x) \sin x}{1+\cos^2 x}dx$

Using $* \int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$

$\displaystyle I =\int_{0}^{\pi}\dfrac{\pi \sin x}{1+\cos^2 x}dx-\int_{0}^{\pi}\dfrac{x \sin x}{1+\cos^2 x}dx$

$\displaystyle I =\int_{0}^{\pi}\dfrac{\pi \sin x}{1+\cos^2 x}dx-I\Leftrightarrow 2I=\pi.\int_{0}^{\pi}\dfrac{\sin x}{1+\cos^2 x}dx$

$\displaystyle I=\dfrac{\pi}{2}.\int_{0}^{\pi}\dfrac{\sin x}{1+\cos^2 x}dx$

Let $\cos x=t\Leftrightarrow -\sin xdx = dt$ and changing the limit

$I=-\dfrac{\pi}{2}\int_{1}^{-1}\dfrac{1}{1+t^2}dt=\dfrac{\pi}{2}\int_{-1}^{1}\dfrac{1}{1+t^2}dt=2\times \dfrac{\pi}{2}\int_{0}^{1}\dfrac{1}{1+t^2}dt$

Using $*\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$

Using $*\int_{-a}^{a}f(x)dx=2.\int_{0}^{a}f(x)dx$ (even function i.e $f(-x)=f(x))$

$\displaystyle I=\pi.\int_{0}^{1}\dfrac{1}{1+t^2}=\pi.\tan^{-1}(t)\Big|_{0}^{1}=\pi.\left\{\tan^{-1}(1)-\tan^{-1}(0)\right\}$

$\displaystyle I=\pi.\dfrac{\pi}{4}=\dfrac{\pi^2}{4}$

[stuart clark]

(a) $\displaystyle\int\dfrac{1}{\sqrt{x^2+k}}dx$

$\dispiaystyle x^2+k=y^2\Leftrightarrow 2xdx=2ydy\Leftrightarrow xdx=ydy$

$\displaystyle \dfrac{dx}{y}=\dfrac{dy}{x}=\dfrac{d(x+y)}{(x+y)}$

Using $* \displaystyle \dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}$

$\displaystyle\int\dfrac{1}{\sqrt{x^2+k}}dx=\int\dfrac{dx}{y}=\dfrac{d(x+y)}{(x+y)}=\ln\left|x+y\right|+C$

$\displaystyle\int\dfrac{1}{\sqrt{x^2+k}}dx=\ln\left|x+\sqrt{x^2+k}\right|+C$


(b) Let $\;\displaystyle{I=\int\sqrt{x^2+k}}.1dx=\sqrt{x^2+k}.x-\int\dfrac{d}{dx}\left(\sqrt{x^2+k}\right).xdx$ (Integration by parts)

$\displaystyle I=x.\sqrt{x^2+k}-\int\dfrac{x^2}{\sqrt{x^2+k}}=x.\sqrt{x^2+k}-\int\dfrac{(x^2+k)-k}{\sqrt{x^2+k}}$

$\displaystyle I=x.\sqrt{x^2+k}-\int\sqrt{x^2+k}dx-k.\int\dfrac{1}{\sqrt{x^2+k}}dx$

$\displaystyle I=x.\sqrt{x^2+k}-I-k.\int\dfrac{1}{\sqrt{x^2+k}}dx$

$\displaystyle 2I=x.\sqrt{x^2+k}-k.\int\dfrac{1}{\sqrt{x^2+k}}dx$

$\displaystyle I=\dfrac{x}{2}.\sqrt{x^2+k}-\dfrac{k}{2}.\int\dfrac{1}{\sqrt{x^2+k}}dx$

$\displaystyle I=\dfrac{x}{2}.\sqrt{x^2+k}-\dfrac{k}{2}.\ln\left|x+\sqrt{x^2+k}\right|+C$

[vietfrog]

Thanks you very much!