7.2: Integrals ya trigonometric

Malengo ya kujifunza

• Kutatua matatizo ya ushirikiano kuwashirikisha bidhaa na mamlaka ya$\sin x$ na$\cos x$.
• Kutatua matatizo ya ushirikiano kuwashirikisha bidhaa na mamlaka ya$\tan x$ na$\sec x$.
• Tumia formula za kupunguza kutatua integrals trigonometric.

Katika sehemu hii tunaangalia jinsi ya kuunganisha bidhaa mbalimbali za kazi za trigonometric. Hizi integrals huitwa integrals trigonometric. Wao ni sehemu muhimu ya mbinu ya ushirikiano iitwayo trigonometric badala, ambayo ni featured katika Trigonometric Badala. Mbinu hii inatuwezesha kubadili maneno ya algebraic ambayo hatuwezi kuunganisha katika maneno yanayohusisha kazi za trigonometric, ambazo tunaweza kuunganisha kwa kutumia mbinu zilizoelezwa katika sehemu hii. Aidha, aina hizi za integrals kuonekana mara kwa mara wakati sisi kujifunza polar, cylindrical, na spherical kuratibu mifumo baadaye. Hebu tuanze utafiti wetu na bidhaa za$\sin x$ na$\cos x.$

Kuunganisha Bidhaa na Mamlaka ya dhambi x na cos x

Wazo muhimu nyuma ya mkakati kutumika kuunganisha mchanganyiko wa bidhaa na mamlaka ya$\sin x$ na$\cos x$ inahusisha kuandika upya maneno haya kama kiasi na tofauti ya integrals ya fomu$∫\sin^jx\cos x\,dx$ au$∫\cos^jx\sin x\,dx$. Baada ya kuandika upya integrals hizi, sisi kutathmini yao kwa kutumia$u$ -badala. Kabla ya kuelezea mchakato wa jumla kwa undani, hebu tuangalie mifano ifuatayo.

Mfano$\PageIndex{1}$: Integrating $\displaystyle ∫\cos^j x\sin x\,dx$

Tathmini$\displaystyle ∫\cos^3x\sin x\,dx.$

Suluhisho

Matumizi$u$ -badala na basi$u=\cos x$. Katika kesi hiyo,$du=−\sin x\,dx.$

Hivyo,

$$∫\cos^3x\sin x\,dx=−∫u^3\,du=−\frac{1}{4}u^4+C=−\frac{1}{4}\cos^4x+C.\nonumber$$

Zoezi$\PageIndex{1}$

Tathmini$\displaystyle ∫\sin^4x\cos x\,dx.$

Kidokezo

Hebu$u=\sin x.$

Jibu

$\displaystyle ∫\sin^4x\cos x\,dx = \frac{1}{5}\sin^5x+C$

Mfano$\PageIndex{2}$: A Preliminary Example: Integrating $∫\cos^jx\sin^kx\,dx$ where $k$ is Odd

Tathmini$\displaystyle ∫\cos^2x\sin^3x\,dx.$

Suluhisho

Kubadili hii muhimu kwa integrals ya$\displaystyle ∫\cos^jx\sin x\,dx,$ fomu upya$\sin^3x=\sin^2x\sin x$ na kufanya badala$\sin^2x=1−\cos^2x.$

Hivyo,

\ (\ displaystyle\ kuanza {align*}\ cos ^ 2x\ dhambi ^ 3x\, dx &=\ cos ^ 2x (1-\ cos ^ 2x)\ dhambi x\, dx & &\ maandishi {Hebu} u=\ cos x;\;\ maandishi {kisha} du=-\ dhambi x\, dx.\\ [4pt]
&=Δ u ^ 2 (1—u ^ 2)\, du\\ [4pt]
&=( u ^ 4,1u ^ 2)\, du\\ [4pt]
&=\ frac {1} {5} u^5—frac {1} {3} U ^ 3+c\\ [4pt]
&=\ frac {1} {5}\ cos^5x\ frac {1} {3}\ COS ^ 3x+C.\ mwisho {align*}\)

Zoezi$\PageIndex{2}$

Tathmini$\displaystyle ∫\cos^3x\sin^2x\,dx.$

Kidokezo

Andika$\cos^3x=\cos^2x\cos x=(1−\sin^2x)\cos x$ na uache$u=\sin x$.

Jibu

$\displaystyle ∫\cos^3x\sin^2x\,dx = \frac{1}{3}\sin^3x−\frac{1}{5}\sin^5x+C$

Katika mfano unaofuata, tunaona mkakati ambayo lazima kutumika wakati kuna tu hata mamlaka ya$\sin x$ na$\cos x$. Kwa integrals ya aina hii, utambulisho

$$\sin^2x=\frac{1}{2}−\frac{1}{2}\cos(2x)=\frac{1−\cos(2x)}{2} \nonumber$$

na

$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)=\frac{1+\cos(2x)}{2} \nonumber$$

ni muhimu sana. Utambulisho huu wakati mwingine hujulikana kama utambulisho wa kupunguza nguvu na wanaweza kuwa inayotokana na utambulisho wa pembe mbili$\cos(2x)=\cos^2x−\sin^2x$ na utambulisho wa Pythagorean$\cos^2x+\sin^2x=1.$

Mfano$\PageIndex{3}$: Integrating an Even Power of $\sin x$

Tathmini$\displaystyle ∫\sin^2x\,dx$.

Suluhisho

Kutathmini hii muhimu, hebu kutumia utambulisho trigonometric$\sin^2x=\frac{1}{2}−\frac{1}{2}\cos(2x).$ Hivyo,

$\displaystyle ∫\sin^2x\,dx=∫\left(\frac{1}{2}−\frac{1}{2}\cos(2x)\right)\,dx=\frac{1}{2}x−\frac{1}{4}\sin(2x)+C.$

Zoezi$\PageIndex{3}$

Tathmini$\displaystyle ∫\cos^2x\,dx.$

Kidokezo

$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$

Jibu

$\displaystyle ∫\cos^2x\,dx = \frac{1}{2}x+\frac{1}{4}\sin(2x)+C$

Mchakato wa jumla wa kuunganisha bidhaa za nguvu za$\sin x$ na$\cos x$ ni muhtasari katika seti zifuatazo za miongozo.

Kutatua matatizo Mkakati: Kuunganisha Bidhaa na Mamlaka ya$\sin x$ and $cos x$

Kuunganisha$\displaystyle \int \cos^jx\sin^kx\,dx$ kutumia mikakati ifuatayo:

1. Kama$k$ ni isiyo ya kawaida, kuandika upya$\sin^kx=\sin^{k−1}x\sin x$ na kutumia utambulisho wa$\sin^2x=1−\cos^2x$ kuandika upya$\sin^{k−1}x$ katika suala la$\cos x$. Unganisha kutumia ubadilishaji$u=\cos x$. Badala hii inafanya$du=−\sin x\,dx.$

2. Kama$j$ ni isiyo ya kawaida, kuandika upya$\cos^jx=\cos^{j−1}x\cos x$ na kutumia utambulisho wa$\cos^2x=1−\sin^2x$ kuandika upya$\cos^{j−1}x$ katika suala la$\sin x$. Unganisha kutumia ubadilishaji$u=\sin x$. Badala hii inafanya$du=\cos x\,dx.$ (Kumbuka: Kama wote wawili$j$ na$k$ ni isiyo ya kawaida, ama mkakati 1 au mkakati 2 inaweza kutumika.)

3. Ikiwa wote wawili$j$ na$k$ ni hata, tumia$\sin^2x=\dfrac{1−\cos(2x)}{2}$ na$\cos^2x=\dfrac{1+\cos(2x)}{2}$. Baada ya kutumia kanuni hizi, kurahisisha na kuomba tena mikakati 1 kwa njia ya 3 kama inafaa.

Mfano$\PageIndex{4}$: Integrating $∫\cos^jx\sin^kx\,dx$ where $k$ is Odd

Tathmini$\displaystyle ∫\cos^8x\sin^5x\,dx.$

Suluhisho

Kwa kuwa nguvu juu ya$\sin x$ ni isiyo ya kawaida, kutumia mkakati 1. Hivyo,

\ (\ displaystyle\ kuanza {align*}\ cos ^ 8x\ dhambi ^ 5x\, dx &=\ cos ^ 8x\ dhambi ^ 4x\ dhambi x\, dx & &\ maandishi {Kuvunja mbali}\ dhambi x.\\ [4pt]
&=\ cos ^ 8x (\ dhambi ^ 2x) ^2\ dhambi x\, dx &\ maandishi {Andika upya}\ dhambi ^ 4x= (\ dhambi ^ 2x) ^2.\\ [4pt]
&=\ cos^8x (1-\ cos ^ 2x) ^2\ dhambi x\, dx & &\ maandishi { Mbadala}\ dhambi ^ 2x=1-\ cos^2x.\\ [4pt]
&=u ^ 8 (1—u ^ 2) ^2 (-du) & &\ maandishi {Hebu} u=\ cos x\ maandishi {na} du=\ dhambi x\, dx.\\ [4pt]
&=( -u ^ 8+2u^ {10} -u^ 12 {}) du & &\ maandishi {Panua.}\ [4pt]
&=Δ\ Frac {1} {9} u ^ 9+\ frac {2} {11} u^ {11} U^ {11} -\ frac {1} {13} u^ {13} +C & & ;\ Nakala {Tathmini muhimu.}\\ [4pt]
&=\ Frac {1} {9}\ cos^9x+\ Frac {2} {11}\ cos^ {11} x\ frac {1} {13}\ cos^ {13} X+C & &\ maandishi {mbadala} u=\ cos x.\ mwisho {align*}\)

Mfano$\PageIndex{5}$: Integrating $∫\cos^jx\sin^kx\,dx$ where $k$ and $j$ are Even

Tathmini$\displaystyle ∫\sin^4x\,dx.$

Solution: Kwa kuwa nguvu juu ya$\sin x$ ni hata$(k=4)$ na nguvu juu$\cos x$ ya hata ni$(j=0),$ lazima kutumia mkakati 3. Hivyo,

\ (\ kuanza {align*}\ displaystyle\ dhambi ^ 4x\, dx &=\ kushoto (\ dhambi ^ 2x\ haki) ^2\, dx & &\ maandishi {Andika upya}\ dhambi ^ 4x=\ kushoto (\ dhambi ^ 2x\ haki) ^2.\\ [4pt]
&=\ kushoto (\ frac {1} {2}} {1} {2}\ cos (2x)\ haki) ^2\, dx & &\ maandishi {mbadala}\ dhambi ^ 2x=\ frac {1} {2} -\ Frac {1} {1} {1} {2}\ cos (2x).\\ [4pt]
& =\ kushoto (\ frac {1} {4} -\ frac {1} {2}\ cos (2x) +\ frac {1} {4}\ cos ^ 2 (2x)\ haki)\, dx & &\ maandishi {Panua}\ kushoto (\ frac {1} {2}\ cos (2x)\ haki) ^2.\\ [4pt]
&=\ kushoto (\ frac {1} {4}} -\ Frac {1} {2}\ cos (2x) +\ frac {1} {4}\ kushoto (\ frac {1} {2} +\ Frac {1} {2}\ cos (4x)\ haki)\, dx & &\ maandishi Tangu {1} {2}\ cos (4x)\ haki)\, dx & &\ maandishi Tangu {}\ cos^2 (2x )\ maandishi {ina nguvu hata, mbadala}\ cos ^2 (2x) =\ Frac {1} {2} +\ Frac {1} {2}\ cos (4x).\\ [4pt]
&=\ kushoto (\ frac {3} {8}\ frac {8} {1} {1} {2}\ cos (2x) +\ frac {1} {8} cos (4x)\ haki)\, dx & &\ maandishi {Kurahisisha.}\\ [4pt]
&=\ Frac {3} {8} x\ Frac {1} {4}\ dhambi (2x) +\ frac {1} {32}\ dhambi (4x) +C &\ maandishi { Tathmini muhimu.}\\ [4pt]\ mwisho {align*}\)

Zoezi$\PageIndex{4}$

Tathmini$\displaystyle ∫\cos^3x\,dx.$

Kidokezo

Matumizi mkakati 2. Andika$\cos^3x=\cos^2x\cos x$ na ubadilishaji$\cos^2x=1−\sin^2x.$

Jibu

$\displaystyle ∫\cos^3x\,dx = \sin x−\frac{1}{3}\sin^3x+C$

Zoezi$\PageIndex{5}$

Tathmini$\displaystyle ∫\cos^2(3x)\,dx.$

Kidokezo

Matumizi mkakati 3. Mbadala$\cos^2(3x)=\frac{1}{2}+\frac{1}{2}\cos(6x)$

Jibu

$\displaystyle ∫\cos^2(3x)\,dx = \frac{1}{2}x+\frac{1}{12}\sin(6x)+C$

Katika baadhi ya maeneo ya fizikia, kama vile quantum mechanics, usindikaji signal, na hesabu ya mfululizo Fourier, mara nyingi ni muhimu kuunganisha bidhaa ambazo ni pamoja$sin(ax), sin(bx), cos(ax),$ na na Integrals$cos(bx).$ hizi ni tathmini kwa kutumia utambulisho trigonometric, kama ilivyoainishwa katika utawala zifuatazo.

Kanuni: Kuunganisha Bidhaa za Sines na Cosines ya Angles tofauti

Kuunganisha bidhaa kuwashirikisha$\sin(ax), \,\sin(bx), \,\cos(ax),$ na$\cos(bx),$ kutumia substitutions

$$\sin(ax)\sin(bx)=\frac{1}{2}\cos((a−b)x)−\frac{1}{2}\cos((a+b)x) \nonumber$$

$$\sin(ax)\cos(bx)=\frac{1}{2}\sin((a−b)x)+\frac{1}{2}\sin((a+b)x) \nonumber$$

$$\cos(ax)\cos(bx)=\frac{1}{2}\cos((a−b)x)+\frac{1}{2}\cos((a+b)x) \nonumber$$

Hizi formula inaweza kuwa inayotokana na jumla ya-angle formula kwa sine na cosine.

Mfano$\PageIndex{6}$: Evaluating $∫\sin(ax)\cos(bx)\,dx$

Tathmini$\displaystyle ∫\sin(5x)\cos(3x)\,dx.$

Solution: Tumia utambulisho$\sin(5x)\cos(3x)=\frac{1}{2}\sin(2x)+\frac{1}{2}\sin(8x).$ Hivyo,

$\displaystyle ∫\sin(5x)\cos(3x)\,dx=∫\frac{1}{2}\sin(2x)+\frac{1}{2}\sin(8x)\,dx=−\frac{1}{4}\cos(2x)−\frac{1}{16}\cos(8x)+C.$

Zoezi$\PageIndex{6}$

Tathmini$\displaystyle ∫\cos(6x)\cos(5x)\,dx.$

Kidokezo

Mbadala$\cos(6x)\cos(5x)=\frac{1}{2}\cos x+\frac{1}{2}\cos(11x).$

Jibu

$\displaystyle ∫\cos(6x)\cos(5x)\,dx = \frac{1}{2}\sin x+\frac{1}{22}\sin(11x)+C$

Kuunganisha Bidhaa na Mamlaka ya$\tan x$ na$\sec x$

Kabla ya kujadili ushirikiano wa bidhaa$\tan x$ na mamlaka ya na$\sec x$, ni muhimu kukumbuka integrals kuwashirikisha$\tan x$ na$\sec x$ sisi tayari kujifunza:

1. $\displaystyle ∫\sec^2x\,dx=\tan x+C$

2. $\displaystyle ∫\sec x\tan x\,dx=\sec x+C$

3. $\displaystyle ∫\tan x\,dx=\ln|\sec x|+C$

4. $\displaystyle ∫\sec x\,dx=\ln|\sec x+\tan x|+C.$

Kwa integrals zaidi ya bidhaa$\tan x$ na nguvu ya na$\sec x$, sisi kuandika tena kujieleza tunataka kuunganisha kama jumla au tofauti ya integrals ya fomu$\displaystyle ∫\tan^jx\sec^2x\,dx$ au$\displaystyle ∫\sec^jx\tan x\,dx$. Kama tunavyoona katika mfano unaofuata, tunaweza kutathmini integrals hizi mpya kwa kutumia u-badala.

Mfano$\PageIndex{7}$: Evaluating $∫\sec^jx\tan x\,dx$

Tathmini$\displaystyle ∫\sec^5 x\tan x\,dx.$

Suluhisho: Anza kwa kuandika upya$\sec^5 x\tan x$ kama$\sec^4 x\sec x\tan x.$

\ (\ displaystyle\ kuanza {align*}\ sec ^ 5x\ tan x\, dx &=\ sec ^ 4 x\ sec x\ tan x\, dx\\ [4pt]
&=u ^ 4\, du & &\ maandishi {Hebu} u=\ sec x\,\ sec x\ tan x\,\ maandishi {kisha},\, du=\ sec x\ tan x\, dx.\\ [4pt]
&=\ tfrac {1} {5} U ^ 5+C & &\ maandishi {Tathmini muhimu.}\\ [4pt]
&=\\ tfrac {1} {5}\ sec ^ 5 x+C &\ maandishi {mbadala}\ sec x = u.\ mwisho {align*}\)

Unaweza kusoma baadhi ya habari ya kuvutia katika tovuti hii kujifunza kuhusu muhimu ya kawaida kuwashirikisha secant.

Zoezi$\PageIndex{7}$

Tathmini$\displaystyle ∫\tan^5x\sec^2x\,dx.$

Kidokezo

Hebu$u=\tan x$ na$du=\sec^2 x.$

Jibu

$\displaystyle ∫\tan^5x\sec^2x\,dx = \tfrac{1}{6}\tan^6x+C$

Sasa tunaangalia mikakati mbalimbali ya kuunganisha bidhaa na nguvu za$\sec x$ na$\tan x.$

Mkakati wa Kutatua matatizo: Kuunganisha$\displaystyle ∫\tan^kx\sec^jx\,dx$

Kuunganisha$\displaystyle ∫\tan^kx\sec^jx\,dx,$ kutumia mikakati ifuatayo:

1. Kama$j$ ni hata na$j≥2,$ kuandika upya$\sec^jx=\sec^{j−2}x\sec^2x$ na$\sec^2x=\tan^2x+1$ kutumia kuandika upya$\sec^{j−2}x$ katika suala la$\tan x$. Hebu$u=\tan x$ na$du=\sec^2x.$

2. Kama$k$ ni isiyo ya kawaida na$j≥1$, kuandika upya$\tan^kx\sec^jx=\tan^{k−1}x\sec^{j−1}x\sec x\tan x$ na$\tan^2x=\sec^2x−1$ kutumia kuandika upya$\tan^{k−1}x$ katika suala la$\sec x$. Hebu$u=\sec x$ na$du=\sec x\tan x\,dx.$ (Kumbuka: Kama$j$ ni hata na$k$ ni isiyo ya kawaida, basi ama mkakati 1 au mkakati 2 inaweza kutumika.)

3. Kama$k$ ni isiyo ya kawaida ambapo$k≥3$ na$j=0$, kuandika upya$\tan^kx=\tan^{k−2}x\tan^2x=\tan^{k−2}x(\sec^2x−1)=\tan^{k−2}x\sec^2x−\tan^{k−2}x.$ Inaweza kuwa muhimu kurudia mchakato huu juu ya$\tan^{k−2}x$ muda.

4. Ikiwa$k$ ni hata na isiyo$j$ ya kawaida, kisha utumie$\tan^2x=\sec^2x−1$ kuelezea$\tan^kx$ kwa suala la$\sec x$. Matumizi ya ushirikiano na sehemu ya kuunganisha nguvu isiyo ya kawaida ya$\sec x.$

Mfano$\PageIndex{8}$: Integrating $∫\tan^kx\sec^jx\,dx$ when $j$ is Even

Tathmini$\displaystyle ∫\tan^6x\sec^4x\,dx.$

Suluhisho

Kwa kuwa nguvu juu$\sec x$ ni hata, kuandika upya$\sec^4x=\sec^2x\sec^2x$ na$\sec^2x=\tan^2x+1$ kutumia kuandika upya kwanza$\sec^2x$ katika suala la$\tan x.$ Hivyo,

\ (\ kuanza {align*}\ displaystyle\ tan^6x\ sec ^ 4x\, dx &=\ tan ^ 6x (\ tan^2x+1)\ sec ^ 2x\, dx\\ [4pt]
&=u ^ 6 (u ^ 2+1)\, du & &\ maandishi {Hebu} u=\ tan x\ maandishi {na} du=\ sec ^ 2x.\\ [4pt]
&=( u ^ 8+u ^ 6)\, du & &\ maandishi {Panua.}\\ [4pt]
&=\ Frac {1} {9} u ^ 9+\ frac {1} {7} u^ 7+C & &\ maandishi {Tathmini muhimu.}\\ [4pt]
&=\ Frac {1} {9}\ tan ^ 9x+\ frac {1} {7}\ Tan ^ 7x+C. & &\ maandishi {mbadala}\ tan x = u.\ mwisho {align*}\)

Mfano$\PageIndex{9}$: Integrating $∫\tan^kx\sec^jx\,dx$ when $k$ is Odd

Tathmini$\displaystyle ∫\tan^5x\sec^3x\,dx.$

Suluhisho

Kwa kuwa nguvu juu$\tan x$ ni isiyo ya kawaida, kuanza kwa kuandika upya$\tan^5x\sec^3x=\tan^4x\sec^2x\sec x\tan x.$ Hivyo,

\ (\ kuanza {align*}\ displaystyle\ tan^5x\ sec ^ 3x\, dx&=\ tan^4x\ sec ^ 2x\ sec x\ tan x.\\ [4pt]
&=(\ tan^2x) ^2\ sec ^ 2x\ sec x\ tan x\, dx & &\ maandishi {Andika}\ tan ^ 4x =(\ tan^4x = (\ tank x\ ^ 2x) ^2.\\ [4pt]
&=(\ sec ^ 2x-1) ^2\ sec ^ 2x\ sec x\ tan x\, dx & &\ maandishi {Matumizi}\ tan^2x=\ sec ^ 2x-1.\\ [4pt]
&=( u ^ 2,11) ^2u ^ 2du & &\ maandishi {Hebu} u=\ sec x\ maandishi {na} du=\ sec x\ tan x\, dx\\ [4pt]
&=( u ^ 6,12u ^ 4+u ^ 2) du & &\ maandishi {Panua.}\\ [4pt]
&=\ frac {1} {7} u ^ 7\ frac {2} {5} u ^ 5+\ Frac {1} {3} U ^ 3+C &\ maandishi {Unganisha.}\\ [4pt]
&=\ Frac {1 } {7}\ sec ^ 7x—Frac {2} {5}\ sec ^ 5x+\ frac {1} {3}\ sec ^ 3x+C & &\ maandishi {mbadala}\ sec x = u.\ mwisho {align*}\)

Mfano$\PageIndex{10}$: Integrating $∫\tan^kx\,dx$ where $k$ is Odd and $k≥3$

Tathmini$\displaystyle ∫\tan^3x\,dx.$

Suluhisho

Anza kwa kuandika upya$\tan^3x=\tan x\tan^2x=\tan x(\sec^2x−1)=\tan x\sec^2x−\tan x.$ Hivyo,

\ ({2}\ kuanza {align*}\ displaystyle\ tan ^ 3x\, dx &= (\ tan x\ sec ^ 2x—tan x)\, dx\\ [4pt]
&=\ tan x\ sec ^ 2x\, dx\\ tan x\, dx\\ [4pt]
&=\ Frac {1}\ tan ^ 2x2x\ -\ ln |\ sec x|+C.\ mwisho {align*}\)

Kwa muhimu ya kwanza, tumia badala$u=\tan x.$ Kwa muhimu ya pili, tumia formula.

Mfano$\PageIndex{11}$: Integrating $\displaystyle ∫\sec^3x\,dx$

Unganisha$\displaystyle ∫\sec^3x\,dx.$

Suluhisho

Hii muhimu inahitaji ushirikiano na sehemu. Kuanza, basi$u=\sec x$ na$dv=\sec^2x$. Uchaguzi huu kufanya$du=\sec x\tan x$ na$v=\tan x$. Hivyo,

\ (\ kuanza {align*}\ displaystyle\ sec ^ 3x\, dx &=\ sec x\ tan x-\ tan x\ sec x\ tan x\\ tan x\, dx\\ [4pt]
&=\ sec x\ tan ^ 2x\ sec x\\, dx & &\ maandishi {Kurahisisha.}\\ [4pt]
&=\ sec x\ tan x-(\ sec ^ 2x-1)\ sec x\, dx & &\ maandishi {mbadala}\ tan^2x=\ sec ^ 2x-1.\\ [4pt]
& amp; =\ sec x\ tan x+\ sec x\, dx-\ sec ^ 3x\, dx & &\ maandishi {Andika upya.}\\ [4pt]
&=\ sec x\ tan x+\ ln|\ sec x+\ tan x|-\ sec ^ 3x\, dx. & &\ maandishi {Tathmini}\ sec x\, dx. \ mwisho {align*}\)

Sasa tuna

$$∫\sec^3x\,dx=\sec x\tan x+\ln|\sec x+\tan x|−∫\sec^3x\,dx.\nonumber$$

Kwa kuwa muhimu$\displaystyle ∫\sec^3x\,dx$ imetokea tena upande wa kulia, tunaweza kutatua kwa kuiongeza$\displaystyle ∫\sec^3x\,dx$ kwa pande zote mbili. Kwa kufanya hivyo, sisi kupata

$$2∫\sec^3x\,dx=\sec x\tan x+\ln|\sec x+\tan x|.\nonumber$$

Kugawanya na 2, tunawasili

$$∫\sec^3x\,dx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x+\tan x|+C\nonumber$$

Zoezi$\PageIndex{8}$

Tathmini$\displaystyle ∫\tan^3x\sec^7x\,dx.$

Kidokezo

Tumia Mfano$\PageIndex{9}$ kama mwongozo.

Jibu

$\displaystyle ∫\tan^3x\sec^7x\,dx = \frac{1}{9}\sec^9x−\frac{1}{7}\sec^7x+C$

Kupunguza formula

Kutathmini maadili$\displaystyle ∫\sec^nx\,dx$ ya$n$ ambapo$n$ ni isiyo ya kawaida inahitaji ushirikiano na sehemu. Aidha, ni lazima pia kujua thamani ya$\displaystyle ∫\sec^{n−2}x\,dx$ kutathmini$\displaystyle ∫\sec^nx\,dx$. Tathmini ya$\displaystyle ∫\tan^nx\,dx$ pia inahitaji kuwa na uwezo wa kuunganisha$\displaystyle ∫\tan^{n−2}x\,dx$. Ili kufanya mchakato iwe rahisi, tunaweza kupata na kutumia kanuni zifuatazo za kupunguza nguvu. Sheria hizi zinatuwezesha kuchukua nafasi muhimu ya nguvu ya$\sec x$ au$\tan x$ kwa umuhimu wa nguvu ya chini ya$\sec x$ au$\tan x.$

Kanuni: Kupunguza formula kwa$∫\sec^nx\,dx$ and $∫\tan^nx\,dx$

$$∫\sec^n x\,dx=\frac{1}{n−1}\sec^{n−2}x\tan x+\frac{n−2}{n−1}∫\sec^{n−2}x\,dx \nonumber$$

$$∫\tan^n x\,dx=\frac{1}{n−1}\tan^{n−1}x−∫\tan^{n−2}x\,dx \nonumber$$

Utawala wa kwanza wa kupunguza nguvu unaweza kuthibitishwa kwa kutumia ushirikiano na sehemu. pili inaweza kuthibitishwa kwa kufuata mkakati ilivyoainishwa kwa ajili ya kuunganisha nguvu isiyo ya kawaida ya$\tan x.$

Mfano$\PageIndex{12}$: Revisiting $∫\sec^3x\,dx$

Tumia formula ya kupunguza ili kutathmini$\displaystyle ∫\sec^3x\,dx.$

Suluhisho: Kwa kutumia formula ya kwanza ya kupunguza, tunapata

\ (\ kuanza {align*}\ displaystyle\ sec ^ 3x\, dx &=\ Frac {1} {2}\ sec x\ tan x+\ frac {1} {2}\ sec x\\, dx\\ [4pt]
&=\ frac {1} {2}\ frac {1}\ ln|\ sekunde x+\ tan x|+C.\ mwisho {align*}\)

Mfano$\PageIndex{13}$: Using a Reduction Formula

Tathmini$\displaystyle ∫\tan^4x\,dx.$

Suluhisho: Kutumia formula ya kupunguza kwa$∫\tan^4x\,dx$ tuna

\ (\ kuanza {align*}\ displaystyle\ tan^4x\, dx &=\ Frac {1} {3}\ tan ^ 3x-\ tan^2x\, dx\\ [4pt]
&=\ frac {1} {3}\ tan ^ 3x\ tan ^ 0x\, dx) & &\ maandishi {formula ya kupunguza kwa}\ tan^2x\, dx.\\ [4pt]
&=\ frac {1} {3}\ tan^3x-tan x+1\, dx & &\ maandishi {Kurahisisha.}\\ [4pt]
&=\ frac {1} {3}\ tan^3x-tan X+X+C &\ maandishi {Tathmini} 1\, dx\ mwisho {align*}\)

Zoezi$\PageIndex{9}$

Tumia formula ya kupunguza$\displaystyle ∫\sec^5x\,dx.$

Kidokezo

Tumia formula ya kupunguza 1 na uache$n=5.$

Jibu

$\displaystyle ∫\sec^5x\,dx=\frac{1}{4}\sec^3x\tan x+\frac{3}{4}∫\sec^3x$

Dhana muhimu

Integrals ya kazi trigonometric inaweza kupimwa na matumizi ya mikakati mbalimbali. Mikakati hii ni pamoja na

1. Kutumia utambulisho wa trigonometric kuandika upya muhimu ili iweze kutathminiwa na$u$ -badala
2. Kutumia ushirikiano na sehemu
3. Kutumia utambulisho wa trigonometric kuandika upya bidhaa za sines na cosines na hoja tofauti kama jumla ya kazi za sine na cosine binafsi
4. Kutumia formula za kupunguza

Mlinganyo muhimu

Kuunganisha bidhaa zinazohusisha$\sin(ax), \,\sin(bx), \,\cos(ax),$ na$\cos(bx),$ kutumia mbadala.

• Bidhaa za Sine

$\sin(ax)\sin(bx)=\frac{1}{2}\cos((a−b)x)−\frac{1}{2}\cos((a+b)x)$

• Bidhaa za Sine na Cosine

$\sin(ax)\cos(bx)=\frac{1}{2}\sin((a−b)x)+\frac{1}{2}\sin((a+b)x)$

• Bidhaa za Cosine

$\cos(ax)\cos(bx)=\frac{1}{2}\cos((a−b)x)+\frac{1}{2}\cos((a+b)x)$

• Mfumo wa Kupunguza Nguvu

$\displaystyle ∫\sec^nx\,dx=\frac{1}{n−1}\sec^{n−2}x \tan x+\frac{n−2}{n−1}∫\sec^{n−2}x\,dx$

• Mfumo wa Kupunguza Nguvu

$\displaystyle ∫\tan^nx\,dx=\frac{1}{n−1}\tan^{n−1}x−∫\tan^{n−2}x\,dx$

faharasa

formula ya kupunguza nguvu

sheria ambayo inaruhusu muhimu ya nguvu ya kazi ya trigonometric kubadilishana kwa muhimu inayohusisha nguvu ya chini

trigonometric muhimu

muhimu kuwashirikisha nguvu na bidhaa za kazi trigonometric