5.7: Integrals Kusababisha Inverse Trigonometric Kazi

Malengo ya kujifunza

• Unganisha kazi zinazosababisha kazi za trigonometric inverse

Katika sehemu hii sisi kuzingatia integrals kwamba kusababisha inverse trigonometric kazi. Tumefanya kazi na kazi hizi kabla. Kumbuka, kwamba kazi trigonometric si moja kwa moja isipokuwa domains ni vikwazo. Wakati wa kufanya kazi na inverses ya kazi za trigonometric, sisi daima tunahitaji kuwa makini kuzingatia vikwazo hivi. Pia, sisi hapo awali tulianzisha fomu za derivatives ya kazi za trigonometric inverse. Fomu zilizotengenezwa huko zinatoa kupanda moja kwa moja kwa formula za ushirikiano zinazohusisha kazi za trigonometric inverse

Integrals kwamba Matokeo katika Inverse Trigonometric Kazi

Hebu tuanze sehemu hii ya mwisho ya sura na kanuni tatu. Pamoja na kanuni hizi, tunatumia badala ya kutathmini integrals. Sisi kuthibitisha formula kwa inverse sine muhimu.

Kanuni: Fomu za Ushirikiano Kusababisha Kazi za Trigonometric Inverse

Fomu zifuatazo za ushirikiano hutoa kazi za trigonometric inverse:

\[ \begin{align} ∫\dfrac{du}{\sqrt{a^2−u^2}} =\sin^{−1}\left(\dfrac{u}{a}\right)+C \\ ∫\dfrac{du}{a^2+u^2} =\dfrac{1}{a}\tan^{−1}\left(\dfrac{u}{a}\right)+C \\ ∫\dfrac{du}{u\sqrt{u^2−a^2}} =\dfrac{1}{a}\sec^{−1}\left(\dfrac{|u|}{a}\right)+C \end{align} \nonumber \]

Uthibitisho wa formula ya kwanza

Hebu\( y=\sin^{−1}\frac{x}{a}\). Kisha\( a \sin y=x\). Sasa kwa kutumia tofauti thabiti, tunapata

\[ \dfrac{d}{dx}(a \sin y)=\dfrac{d}{dx}(x) \nonumber \]

\[ a\cos y\dfrac{dy}{dx}=1 \nonumber \]

\[ \dfrac{dy}{dx}=\dfrac{1}{a\cos y}. \nonumber \]

Kwa\( −\dfrac{π}{2}≤y≤\dfrac{π}{2},\cos y≥0.\) hiyo, kutumia utambulisho wa Pythagorean\( \sin^2y+\cos^2y=1\), tuna\( \cos y=\sqrt{1-\sin^2y}.\) Hii inatoa

\[ \begin{align} \dfrac{1}{a \cos y} =\dfrac{1}{a\sqrt{1−\sin^2y}} \\ =\dfrac{1}{\sqrt{a^2−a^2 \sin^2y}} \\ =\dfrac{1}{\sqrt{a^2−x^2}}. \end{align} \nonumber \]

Basi kwa ajili ya\( −a≤x≤a,\) tuna

\[ ∫\dfrac{1}{\sqrt{a^2−u^2}}\,du=\sin^{−1}\left(\frac{u}{a}\right)+C. \nonumber \]

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Mfano\( \PageIndex{1}\): Evaluating a Definite Integral Using Inverse Trigonometric Functions

Tathmini muhimu ya uhakika

\[ ∫^{1/2}_0\dfrac{dx}{\sqrt{1−x^2}}. \nonumber \]

Suluhisho

Tunaweza kwenda moja kwa moja kwa formula kwa antiderivative katika utawala juu ya formula ushirikiano kusababisha inverse trigonometric kazi, na kisha kutathmini muhimu uhakika. Tuna

\[\int_0^{1/2}\dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1} x \,\bigg|_0^{1/2} = \sin^{-1} \tfrac{1}{2} - \sin^{-1} 0 = \dfrac{\pi}{6}-0 = \dfrac{\pi}{6}. \nonumber \]

Kumbuka kuwa tangu integrand ni tu derivative ya\(\sin^{-1} x\), sisi ni kweli tu kutumia ukweli huu kupata antiderivative hapa.

Zoezi\(\PageIndex{1}\)

Kupata muhimu kwa muda usiojulikana kwa kutumia inverse trigonometric kazi na badala ya\(\displaystyle ∫\dfrac{dx}{\sqrt{9−x^2}}\).

Kidokezo

Tumia formula katika utawala juu ya formula za ushirikiano na kusababisha kazi za trigonometric inverse.

Jibu

\( \displaystyle ∫\dfrac{dx}{\sqrt{9−x^2}} \quad=\quad \sin^{−1}\left(\dfrac{x}{3}\right)+C \)

Katika integrals nyingi zinazosababisha kazi za trigonometric inverse katika antiderivative, tunaweza kuhitaji kutumia badala ili kuona jinsi ya kutumia formula za ushirikiano zilizotolewa hapo juu.

Mfano\( \PageIndex{2}\): Finding an Antiderivative Involving an Inverse Trigonometric Function using substitution

Tathmini muhimu

\[ ∫\dfrac{dx}{\sqrt{4−9x^2}}.\nonumber \]

Suluhisho

Mbadala\( u=3x\). Kisha\( du=3\,dx\) na tuna

\[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}.\nonumber \]

Kutumia formula na\( a=2,\) sisi kupata

\[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}=\dfrac{1}{3}\sin^{−1}\left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\sin^{−1}\left(\dfrac{3x}{2}\right)+C.\nonumber \]

Zoezi\(\PageIndex{2}\)

Kupata antiderivative ya\(\displaystyle ∫\dfrac{dx}{\sqrt{1−16x^2}}.\)

Kidokezo

Mbadala\( u=4x\).

Jibu

\( \displaystyle ∫\dfrac{dx}{\sqrt{1−16x^2}} = \dfrac{1}{4}\sin^{−1}(4x)+C\)

Mfano\( \PageIndex{3}\): Evaluating a Definite Integral

Tathmini muhimu ya uhakika

\[ ∫^{\sqrt{3}/2}_0\dfrac{du}{\sqrt{1−u^2}}\nonumber. \nonumber \]

Suluhisho

Fomu ya tatizo inafanana na formula ya sine inverse. Hivyo,

\[ ∫^{\sqrt{3}/2}_0\dfrac{du}{\sqrt{1−u^2}}=\sin^{−1}u\,\bigg|^{\sqrt{3}/2}_0=[\sin^{−1}\left(\dfrac{\sqrt{3}}{2}\right)]−[\sin^{−1}(0)]=\dfrac{π}{3}.\nonumber \]

Integrals Kusababisha nyingine Inverse Trigonometric Kazi

Kuna kazi sita za trigonometric inverse. Hata hivyo, formula tatu tu za ushirikiano zinajulikana katika utawala juu ya formula za ushirikiano zinazosababisha kazi za trigonometric inverse kwa sababu tatu zilizobaki ni matoleo hasi ya yale tunayotumia. Tofauti pekee ni kama integrand ni chanya au hasi. Badala ya kukariri formula tatu zaidi, ikiwa integrand ni hasi, tu factor nje -1 na kutathmini muhimu kwa kutumia moja ya formula tayari zinazotolewa. Ili kufunga sehemu hii, tunachunguza formula moja zaidi: muhimu inayosababisha kazi ya tangent inverse.

Mfano\( \PageIndex{4}\): Finding an Antiderivative Involving the Inverse Tangent Function

Kupata antiderivative ya\(\displaystyle ∫\dfrac{1}{9+x^2}\,dx.\)

Suluhisho

Tumia formula na\( a=3\). Kisha,

\[ ∫\dfrac{dx}{9+x^2}=\dfrac{1}{3}\tan^{−1}\left(\dfrac{x}{3}\right)+C. \nonumber \]

Zoezi\(\PageIndex{3}\)

Kupata antiderivative ya\(\displaystyle ∫\dfrac{dx}{16+x^2}\).

Kidokezo

Fuata hatua katika Mfano\( \PageIndex{4}\).

Jibu

\(\displaystyle ∫\dfrac{dx}{16+x^2} = \frac{1}{4}\tan^{−1}\left(\dfrac{x}{4}\right)+C \)

Mfano\( \PageIndex{5}\): Applying the Integration Formulas WITH SUBSTITUTION

Kupata antiderivative ya\(\displaystyle ∫\dfrac{1}{1+4x^2}\,dx.\)

Suluhisho

Kulinganisha tatizo hili na kanuni zilizotajwa katika utawala juu ya formula za ushirikiano zinazosababisha kazi za trigonometric inverse, integrand inaonekana sawa na formula kwa\( \tan^{−1} u+C\). Hivyo sisi kutumia badala, kuruhusu\( u=2x\), basi\( du=2\,dx\) na\( \dfrac{1}{2}\,du=dx.\) kisha, tuna

\[ \dfrac{1}{2}∫\dfrac{1}{1+u^2}\,du=\dfrac{1}{2}\tan^{−1}u+C=\dfrac{1}{2}\tan^{−1}(2x)+C. \nonumber \]

Zoezi\(\PageIndex{4}\)

Matumizi badala ya kupata antiderivative ya\(\displaystyle ∫\dfrac{dx}{25+4x^2}.\)

Kidokezo

Tumia mkakati wa kutatua kutoka Mfano\( \PageIndex{5}\) na utawala juu ya formula za ushirikiano na kusababisha kazi inverse trigonometric.

Jibu

\(\displaystyle ∫\dfrac{dx}{25+4x^2} = \dfrac{1}{10}\tan^{−1}\left(\dfrac{2x}{5}\right)+C \)

Mfano\( \PageIndex{6}\): Evaluating a Definite Integral

Tathmini muhimu ya uhakika\(\displaystyle ∫^{\sqrt{3}}_{\sqrt{3}/3}\dfrac{dx}{1+x^2}\).

Suluhisho

Tumia formula kwa tangent inverse. Tuna

\[\int_{\sqrt{3}/3}^{\sqrt{3}} \frac{dx}{1+x^2} = \tan^{-1}x\,\bigg|_{\sqrt{3}/3}^{\sqrt{3}} = [\tan^{-1}\left(\sqrt{3}\right)] - [\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)] =\frac{\pi}{3} - \frac{\pi}{6} =\frac{\pi}{6}.\nonumber \]

Zoezi\(\PageIndex{5}\)

Tathmini muhimu ya uhakika\(\displaystyle ∫^2_0\dfrac{dx}{4+x^2}\).

Kidokezo

Fuata taratibu kutoka Mfano\(\PageIndex{6}\) ili kutatua tatizo.

Jibu

\(\displaystyle ∫^2_0\dfrac{dx}{4+x^2} = \dfrac{π}{8} \)

Dhana muhimu

• Fomula kwa derivatives ya kazi inverse trigonometric maendeleo katika Derivatives ya Kielelezo na Logarithmic Kazi kusababisha moja kwa moja kwa formula ushirikiano kuwashirikisha inverse trigon
• Tumia fomu zilizoorodheshwa katika utawala juu ya formula za ushirikiano na kusababisha kazi za trigonometric inverse ili kufanana na muundo sahihi na kufanya mabadiliko kama inavyohitajika ili kutatua tatizo.
• Kubadilishwa mara nyingi inahitajika kuweka integrand katika fomu sahihi.

Mlinganyo muhimu

• Integrals kwamba kuzalisha Inverse Trigonometric Kazi

\(\displaystyle ∫\dfrac{du}{\sqrt{a^2−u^2}}=\sin^{−1}\left(\dfrac{u}{a}\right)+C\)

\(\displaystyle ∫\dfrac{du}{a^2+u^2}=\dfrac{1}{a}\tan^{−1}\left(\dfrac{u}{a}\right)+C\)

\(\displaystyle ∫\dfrac{du}{u\sqrt{u^2−a^2}}=\dfrac{1}{a}\sec^{−1}\left(\dfrac{|u|}{a}\right)+C\)

Wachangiaji na Majina

• Template:ContribOpenStaxCalc
• Includes some added textual clarifications and edits by Paul Seeburger (Monroe Community College)