Funktion $$$- \tan{\left(x \right)} + \sec{\left(x \right)}$$$ integraali

Määritä $$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)\, dx$$$.

Integroi termi kerrallaan:

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

Kirjoita tangentti uudelleen muotoon $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

$$\int{\sec{\left(x \right)} d x} - {\color{red}{\int{\tan{\left(x \right)} d x}}} = \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

Olkoon $$$u=\cos{\left(x \right)}$$$.

Tällöin $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$\sin{\left(x \right)} dx = - du$$$.

Näin ollen,

$$\int{\sec{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=-1$$$ ja $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$\int{\sec{\left(x \right)} d x} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = \int{\sec{\left(x \right)} d x} - {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$

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

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

Muista, että $$$u=\cos{\left(x \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\sec{\left(x \right)} d x} = \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)} + \int{\sec{\left(x \right)} d x}$$

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

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\sec{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}$$

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:

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}$$

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

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

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

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

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

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{u} d u}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

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

Näin ollen,

$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}$$

Lisää integrointivakio:

$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}+C$$

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