Funktion $$$\sec^{5}{\left(x \right)}$$$ integraali

Laskin löytää funktion $$$\sec^{5}{\left(x \right)}$$$ integraalin/alkufunktion ja näyttää vaiheet.

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Syötteesi

Määritä $$$\int \sec^{5}{\left(x \right)}\, dx$$$.

Ratkaisu

Integraalin $$$\int{\sec^{5}{\left(x \right)} d x}$$$ kohdalla käytä osittaisintegrointia $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Olkoon $$$\operatorname{u}=\sec^{3}{\left(x \right)}$$$ ja $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Tällöin $$$\operatorname{du}=\left(\sec^{3}{\left(x \right)}\right)^{\prime }dx=3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} dx$$$ (vaiheet ovat nähtävissä ») ja $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (vaiheet ovat nähtävissä »).

Integraali voidaan kirjoittaa muotoon

$$\int{\sec^{5}{\left(x \right)} d x}=\sec^{3}{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot 3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - \int{3 \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}$$

Vedä vakio ulos:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - \int{3 \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}$$

Sovella kaavaa $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{3}{\left(x \right)} d x}$$

Laajenna:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{5}{\left(x \right)} - \sec^{3}{\left(x \right)}\right)d x}$$

Erotuksen integraali on integraalien erotus:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{5}{\left(x \right)} - \sec^{3}{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} + 3 \int{\sec^{3}{\left(x \right)} d x} - 3 \int{\sec^{5}{\left(x \right)} d x}$$

Näin saamme seuraavan yksinkertaisen lineaarisen yhtälön integraalin suhteen:

$${\color{red}{\int{\sec^{5}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} + 3 \int{\sec^{3}{\left(x \right)} d x} - 3 {\color{red}{\int{\sec^{5}{\left(x \right)} d x}}}$$

Ratkaisemalla sen saamme, että

$$\int{\sec^{5}{\left(x \right)} d x}=\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \int{\sec^{3}{\left(x \right)} d x}}{4}$$

Integraalin $$$\int{\sec^{3}{\left(x \right)} d x}$$$ kohdalla käytä osittaisintegrointia $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Olkoon $$$\operatorname{u}=\sec{\left(x \right)}$$$ ja $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Tällöin $$$\operatorname{du}=\left(\sec{\left(x \right)}\right)^{\prime }dx=\tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (vaiheet ovat nähtävissä ») ja $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (vaiheet ovat nähtävissä »).

Näin ollen,

$$\int{\sec^{3}{\left(x \right)} d x}=\sec{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot \tan{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}$$

Sovella kaavaa $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}$$

Laajenna:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}$$

Erotuksen integraali on integraalien erotus:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - \int{\sec^{3}{\left(x \right)} d x}$$

Näin saamme seuraavan yksinkertaisen lineaarisen yhtälön integraalin suhteen:

$${\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}$$

Ratkaisemalla sen saamme, että

$$\int{\sec^{3}{\left(x \right)} d x}=\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{\int{\sec{\left(x \right)} d x}}{2}$$

Näin ollen,

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}}{4} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 {\color{red}{\left(\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{\int{\sec{\left(x \right)} d x}}{2}\right)}}}{4}$$

Kirjoita sekantti uudelleen muodossa $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\sec{\left(x \right)} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{8}$$

Kirjoita kosini sinin avulla kaavaa $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ käyttäen ja kirjoita sitten sini uudelleen kaksinkertaisen kulman kaavaa $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$ käyttäen:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8}$$

Kerro osoittaja ja nimittäjä luvulla $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8}$$

Olkoon $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$.

Tällöin $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

Integraali muuttuu muotoon

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

Funktion $$$\frac{1}{u}$$$ integraali on $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

Muista, että $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:

$$\frac{3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} = \frac{3 \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}$$

Näin ollen,

$$\int{\sec^{5}{\left(x \right)} d x} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}$$

Lisää integrointivakio:

$$\int{\sec^{5}{\left(x \right)} d x} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}+C$$

Vastaus

$$$\int \sec^{5}{\left(x \right)}\, dx = \left(\frac{3 \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right)}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}\right) + C$$$A


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