mtk: integral subsitusi trigonometri (1)

$\int \frac{1}{\sqrt{a^2-x^2}}dx$
misalkan :
$x=asint\Rightarrow t=\sin^{-1}\frac{x}{a},-\frac{\Pi }{2}\leq t\leq \frac{\Pi }{2}$
$dx=acostdt$

maka,

$\int \frac{1}{\sqrt{a^2-x^2}}dx=\int \frac{acostdt}{\sqrt{a^2-(asint)^2}}=\int\frac{acostdt}{\sqrt{a^2-a^2sin^2t}}=\int \frac{acostdt}{\sqrt{a^2(1-sin^2t)}}\Rightarrow$

$\int \frac{1}{\sqrt{a^2-x^2}}dx=\int \frac{acostdt}{acost}=\int dt=t+C=\sin^{-1}\frac{x}{a}+C$

$\int \sqrt{a^2-x^2}dx$

misalkan
$x=asint\Rightarrow t=\sin^{-1}\frac{x}{a},-\frac{\Pi }{2}\leq t\leq \frac{\Pi }{2}$
$dx=acostdt$

maka,

$\int \sqrt{a^2-x^2}dx=\int \sqrt{a^2-(asint)^2}(acostdt)=\int \sqrt{a^2-a^2sin^2t}(acostdt)$

$\int \sqrt{a^2-x^2}dx=\int (acost)(acostdt)=a^2\int cos^2tdt\Rightarrow$

$\int \sqrt{a^2-x^2}dx=a^2\int (\frac{1}{2}cos2t+\frac{1}{2})dt=a^2(\frac{1}{4}sin2t+\frac{1}{2}t)+C$

$\int \sqrt{a^2-x^2}dx=\frac{a^2}{4}sin2t+\frac{1}{2}a^2t+C=\frac{a^2}{2}sintcost+\frac{1}{2}a^2t+C$

$\int \sqrt{a^2-x^2}dx=\frac{a^2}{2}sin(\sin^{-1}\frac{x}{a})cos(\sin^{-1}\frac{x}{a})+\frac{1}{2}a^2\sin^{-1}\frac{x}{a}+C$

$\int \sqrt{a^2-x^2}dx=\frac{a^2}{2}.\frac{x}{a}\sqrt{1-(\frac{x}{a})^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C=\frac{ax}{2}\sqrt{\frac{a^2-x^2}{a^2}}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

mtk

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Senin, 03 Januari 2011

integral subsitusi trigonometri (1)

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