\(\int x \cdot \cos(3x) \ \mathrm{d}x\)

\(\text{setze } u=x ,\ \mathrm{d}v= \cos \ 3x \ \mathrm{d}x\)

\(\frac {\mathrm{d}x} {\mathrm{d}x} =\frac {\mathrm{d}u} {\mathrm{d}x} \)   \(\mathrm{d}x=\mathrm{d}u\)

\(v= \frac 1 3 \sin 3x\)

\(\int x \cdot \cos(3x) \ \mathrm{d}x = x\cdot \frac 1 3 \cdot \sin 3x - \int \frac 1 3 \cdot \sin 3x \ \mathrm{d}x= \frac 1 3 \cdot x \cdot \sin 3x - \frac 1 3 \cdot \int \sin 3x \ \mathrm{d}x =\)

\(\frac 1 3 \cdot x \cdot \sin 3x - \frac 1 3 \cdot (-\cos 3x) \cdot \frac 1 3 = \frac 1 3 \cdot x \cdot \sin 3x + \frac 1 9 \cdot \cos 3x +C\)

\(\int x \cdot \sin (7x) \ \mathrm{d}x\)

\(\text{setze } x=u \text{, } \mathrm{d}v=\sin 7x \ \mathrm{d}x\)

\(\frac {\mathrm{d}x}{\mathrm{d}x}=\frac{\mathrm{d}u}{\mathrm{d}x} \)     \(\mathrm{d}x=\mathrm{d}u\)

\(v= - \frac 1 7 \cdot \cos 7x\)

\(\int x \cdot \sin (7x) \ \mathrm{d}x = x \cdot \frac 1 7(-\cos 7x)-\int \frac 1 7 (-cos7x)\mathrm{d} x= \)

\(-\frac 1 7 \cdot x \cdot \cos 7x + \frac 1 7 \cdot \int \cos 7x \ \mathrm{d}x= -\frac 1 7 \cdot x \cdot \cos 7x + \frac 1 7 \cdot \sin 7x \cdot \frac 1 7 +C = \)\(-\frac 1 7 \cdot x \cdot \cos 7x + \frac 1 {49} \cdot \sin 7x +C \)

\(\int x \cdot \cos \frac x 2 \ \mathrm{d}x\)

\(\text {setze} u=x \text{, } \mathrm{d}v= \cos \frac {x}{2} \mathrm{d}x\)

\(\frac{\mathrm{d}x}{\mathrm{d}x}=\frac{\mathrm{d}x}{\mathrm{d}x}\)   \(\mathrm{d}x=\mathrm{d}u\)

\(v= 2\cdot \sin\frac x2\)

\(\int x \cdot \cos \frac x 2 \ \mathrm{d}x = x \cdot 2 \sin \frac x2 \ \mathrm{d}x= 2 x \sin \frac x2 - 2 \cdot \int \sin \frac x2 \ \mathrm{d}x=\)

\(2 x \sin \frac x2 - 2 (-cos\frac x2)2 = 2 x \sin \frac x2 + 4 \cos\frac x2 +C\)

\(\int (5x-4) \cdot \cos (8x) \ \mathrm{d}x\)

\(\text{setze } u=5x-4 \text{, } \mathrm{d}v= \cos (8x) \ \mathrm{d}x\)

\(\frac {\mathrm{d}}{\mathrm{d}x} (5x-4)= \frac{\mathrm{d}u}{\mathrm{d}x}\)    \(\mathrm{d}u=5 \ \mathrm{d}x\)

\(v=\frac 18 \cdot \sin 8x\)

\(\int (5x-4) \cdot \cos (8x) \ \mathrm{d}x= (5x-4)\cdot \frac 18\cdot \sin 8x - \int \frac 18 \sin 8x \cdot 5 \ \mathrm{d}x=\)

\(\frac 18 (5x-4) \sin 8x - \frac 58 \cdot \int \sin 8x \ \mathrm{d}x = \frac 18 (5x-4) \sin 8x - \frac 58 \cdot \frac 18 ( - \cos 8x)+C= \frac 18 (5x-4) \sin 8x + \frac 5{64} \cos 8x +C\)

\(\int (3x+2) \cdot \sin(2x) \ \mathrm{d}x\)

\(\text{setze } u=3x+2 \text{, } \mathrm{d}v=\sin 2x \ \mathrm{d}x\)

\(\frac{\mathrm{d}}{\mathrm{d}x}(3x+2)= \frac{\mathrm{d}u}{\mathrm{d}x}\)   \(\mathrm{d}u=3 \ \mathrm{d}x\)

\(v= -\frac 12 \cos 2x\)

\(\int (3x+2) \cdot \sin(2x) \ \mathrm{d}x= (3x+2)\left( -\frac 12 \cos 2x \right) - \int \left( -\frac 12 \cos 2x \right)3 \ \mathrm{d}x =\)

\(-\frac 12 (3x+2)\cos 2x + \frac 32 \cdot \int \cos 2x \ \mathrm{d}x= -\frac 12 (3x+2)\cos 2x + \frac 34 \cdot \sin 2x + C\)

\(\int (-5x+12) \cdot \cos(3x) \ \mathrm{d}x \)

\(\text{setze } u=-5x+12 \text{, } \mathrm{d}v=\cos 3x \ \mathrm{d}x\)

\(\frac{\mathrm{d}}{\mathrm{d}x}(-5x+12)= \frac{\mathrm{d}u}{\mathrm{d}x}\)    \(\mathrm{d}u=(-5) \ \mathrm{d}x\)

\(v= -\frac 13 \sin 3x\)

\(\int (-5x+12) \cdot \cos(3x) \ \mathrm{d}x= (-5x+12)\frac 13 \sin 3x - \int \frac 13 \sin 2x (-5) \ \mathrm{d}x =\)

\(\frac 13 (-5x+12)\sin 3x - \frac 53 \cdot \frac 13 \cos 3x +C = \frac 13 (-5x+12)\sin 3x - \frac 59 \cos 3x +C\)

\(\text{setze } u=\frac x3 +5 \text{, } \mathrm{d}v=\sin(7x) \ \mathrm{d}x\)

\(\frac{\mathrm{d}}{\mathrm{d}x}(\frac x3 +5)= \frac{\mathrm{d}u}{\mathrm{d}x}\)  \(\mathrm{d}u=\frac 13 \ \mathrm{d}x\)

\(v= -\frac 17 \cos 7x\)

\(\int (\frac x3 +5) \cdot \sin(7x) \ \mathrm{d}x= \left( \frac x3 + 5\right) \left( -\frac 17 \cos 7x\right) -\int \left( -\frac 17 \cos 7x\right) \frac 13 \ \mathrm{d}x =\)

\(-\frac 17 \left( \frac x3 +5\right) \cos 7x + \frac 1{21} \int \cos 7x \ \mathrm{d}x = -\frac 17 \left( \frac x3 +5\right) \cos 7x + \frac 1{147} \sin 7x +C\)

\(\int\limits_{0}^{2\pi} sin^2 nx \ \mathrm{d}x=\)

\(\int\limits_{0}^{2\pi} \frac 12 (1-\cos 2nx)\ \mathrm{d}x= \frac 12 2\pi = \pi\)

\( \left. \int\limits_{0}^{3\tau} A\cdot t\cdot e^{-\frac t\tau} \mathrm{d}t= A \left( t (-\tau) e^{-\frac t\tau} +\tau \cdot \int e^{-\frac t\tau} \mathrm{d}t\right) \right|_{0}^{3\tau}=\)

\(\left. A ( -\tau \cdot t \cdot e^{-\frac t\tau} - \tau^2 \cdot e^{-\frac t\tau} ) \right|_{0}^{3\tau}= A \left[(-3\tau^2 e^{-3} -\tau^2 e^{-3}) -(0-\tau^2) \right]= \)

\(A \tau^2 \left( 1-\frac 4{e^3}\right) \)

\(\text{setze } u=-\frac t\tau\)

\(\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}(-\frac t\tau)\)

\(\mathrm{d}t=-\tau \ \mathrm{d}u\)

\(\int U \cdot e^{-\frac t\tau}\ \mathrm{d}t= \int U \cdot e^{u}\cdot (-\tau)\ \mathrm{d}u=\)

\( U \cdot e^{u}\cdot (-\tau)\ = -U\tau e^{-\frac t\tau}+C\)

\(\text{setze } u=-3x+2\)

\(\frac {\mathrm{d}u}{\mathrm{d}x}= \frac {\mathrm{d}}{\mathrm{d}x} (-3x+2)\)

\(\mathrm{d}u=-3 \ \mathrm{d}x\)

\(\text{setze } \mathrm{d}v=\cos \frac x4 \ \mathrm{d}x\)

\(v=4 \sin \frac x4\)

\(\int (-3x+2)\cdot \cos \frac x4 \ \mathrm{d}x= (-3x+2)\cdot 4 \cdot \sin \frac x4 -\int 4\cdot \sin \frac x4 \cdot (-3) \mathrm{d}x =\)

\(4(-3x+2)\sin \frac x4+12\cdot 4\left( -\cos \frac x4\right) +C= 4(-3x+2)\sin \frac x4 - 48 \cos \frac x4 +C\)

\(\text{setze } u=\left(-\frac{1}{\pi}\cdot x + 1\right) \)

\(\frac {\mathrm{d}u}{\mathrm{d}x}= \frac {\mathrm{d}}{\mathrm{d}x} \left(-\frac{1}{\pi}\cdot x + 1\right)\)

\(\mathrm{d}u=-\frac 1\pi \ \mathrm{d}x\)

\(\text{setze } \mathrm{d}v=\sin nx \ \mathrm{d}x\)

\(v=-\frac 1n \cos nx\)

\(\int \left(-\frac{A}{\pi}\cdot x + A\right) \cdot \sin nx \ \mathrm{d}x= \)

\(A\left[ \left( -\frac 1\pi x +1\right) \frac 1n (-\cos nx)-\int \frac 1n (-\cos nx) \left( -\frac 1\pi\right) \ \mathrm{d}x\right] =\)

\(A\left[ \frac 1n \left( \frac 1\pi x -1\right) \cos nx - \frac 1n \cdot \frac 1\pi \int \cos nx \ \mathrm{d}x\right] =\)

\(A\left[ \frac 1n \left( \frac 1\pi x -1\right) \cos nx - \frac 1n \cdot \frac 1\pi \cdot \frac 1n \sin nx\right] +C=\)

\(A\left[ \frac 1n \left( \frac 1\pi x -1\right) \cos nx - \frac 1{n^2} \cdot \frac 1\pi \cdot \sin nx\right] +C\)

\(\int e^x (2x^2+4) \ \mathrm{d}x \)
 

setze \(f'(x)=e^x\) , \(f(x)=e^x\)und \(g(x)=2x^2+4\), \(g'(x)=4x\)

\(\int e^x (2x^2+4)\ \mathrm{d}x= e^x (2x^2+4) -\int e^x 4x \ \mathrm{d}x\)

noch mal partiell mit \(g(x)=4x \) , \(g'(x)=4\)

\( e^x (2x^2+4) -\int e^x 4x \ \mathrm{d}x = \\ e^x (2x^2+4) - [ e^x 4x -\int e^x 4 \ \mathrm{d}x] =\\ e^x (2x^2+4) - e^x 4x + e^x 4 = e^x(2x^2-4x +8)\)

\(\int e^x (2x^2+4) \ \mathrm{d}x \)
 

setze \(f'(x)=e^x\) , \(f(x)=e^x\)und \(g(x)=2x^2+4\), \(g'(x)=4x\)

\(\int e^x (2x^2+4)\ \mathrm{d}x= e^x (2x^2+4) -\int e^x 4x \ \mathrm{d}x\)

noch mal partiell mit \(g(x)=4x \) , \(g'(x)=4\)

\( e^x (2x^2+4) -\int e^x 4x \ \mathrm{d}x = \\ e^x (2x^2+4) -[ e^x 4x -\int e^x 4 \ \mathrm{d}x ] =\\ e^x (2x^2+4) - e^x 4x + e^x 4 = e^x(2x^2-4x+8)\)

\(\int \limits_0^1 e^x (2x^2+4) \mathrm{d}x= \\ \left. e^x(2x^2-4x+8) \right|_0^1 =\\ e(2-4+8)-1=6e-8 \)

\(\int x \cdot \cos(2x) \ \mathrm{d}x\)

\(\text{setze } g=x ,\ f'= \cos \ 2x\) 

\(g'=1, f=\frac 12 \sin 2x\)

\(\int x \cdot \cos(2x) \ \mathrm{d}x = x\cdot \frac 1 2 \cdot \sin 2x - \int \frac 1 2 \cdot \sin 2x \ \mathrm{d}x= \frac 1 2 \cdot x \cdot \sin 2x - \frac 1 2 \cdot \int \sin 2x \ \mathrm{d}x =\)

\(\frac 1 2 \cdot x \cdot \sin 2x - \frac 1 2 \cdot (-\cos 2x) \cdot \frac 1 2 = \frac 1 2 \cdot x \cdot \sin 2x + \frac 1 4 \cdot \cos 2x\)

\(\left. \frac 1 2 \cdot x \cdot \sin 2x + \frac 1 4 \cdot \cos 2x \right |_{6}^{7}=\\ \)

\(0.5 \cdot 7 \cdot \sin (14) + 0.25 \cdot \cos (14) -(0.5 \cdot 6 \cdot \sin (12) + 0.25 \cdot \cos (12)) \approx\\ 3.5 - (-1.4)=4.9\)

\(\int x \cdot e^x \ \mathrm{d}x\)

\(\text{setze } u=x ,\ \mathrm{d}v= e^x \ \mathrm{d}x\)

\(v= e^x\)

\(\int x \cdot e^x \ \mathrm{d}x = x \cdot e^x - \int e^x \mathrm{d}x = x \cdot e^x - e^x = e^x \cdot (x-1)\)

\(\int x^2 \cdot e^x \ \mathrm{d}x \)

\(\text{setze } f(x)=x^2 ,\ g'(x)= e^x \)

\(f'(x) = 2x \ ; \ g(x) = e^x\)

\(\int x^2 \cdot e^x \ \mathrm{d}x = x^2 \cdot e^x - \int 2 x \cdot e^x \mathrm{d}x \)

nochmal partielle Integration mit

\(\text{setze } f(x)=2x ,\ g'(x)= e^x \)

\(f'(x) = 2 \ ; \ g(x) = e^x\)

\( x^2 \cdot e^x - \int 2 x \cdot e^x \mathrm{d}x = \\ x^2 \cdot e^x - ( 2 x \cdot e^x - \int 2 \cdot e^x \mathrm{d}x ) = \\ x^2 \cdot e^x - ( 2 x \cdot e^x - 2 \cdot e^x ) = e^x ( x^2 - 2x + 2) \)

 \(\int x \cdot \ln (x) \ \mathrm{d}x\)

\(f'(x) = x \ , \ g(x) = \ln(x) \)

\(f(x) = \frac{x^2}{2} \ , \ g'(x) = \frac{1}{x}\)

\(\int x \cdot \ln (x) \ \mathrm{d}x = \frac{x^2}{2} \cdot \ln(x) - \int \frac{1}{x} \cdot \frac{x^2}{2}\mathrm{d}x = \\ \frac{x^2}{2} \cdot \ln(x) - \int \frac{x}{2}\mathrm{d}x = \\ \)

\( \frac{x^2}{2} \cdot \ln(x) - \frac{1}{2} \cdot \frac{x^2}{2}= \\\)

\(\frac{x^2}{2} \cdot (\ln(x) - \frac{1}{2} ) \)

\(\int x \cdot e^{-x} \mathrm{d}x \)

\(setze: f(x) = x \ ; g'(x) = e^{-x}\)

\( f'(x) = 1 \ ; g(x) = - e^{-x}\)

\(\int x \cdot e^{-x} \mathrm{d}x = - x \cdot e^{-x} - \int 1 \cdot (-1) \cdot e^{-x} \mathrm{d}x = \)

\(-x \cdot e^{-x} - e^{-x} = e^{-x} \cdot (-x-1)\)

\(\int x \cdot e^{-x+1} \mathrm{d}x \)

\(setze: f(x) = x \ ; g'(x) = e^{-x+1}\)

\( f'(x) = 1 \ ; g(x) = - e^{-x+1}\)

\(\int x \cdot e^{-x+1} \mathrm{d}x = - x \cdot e^{-x+1} - \int 1 \cdot (-1) \cdot e^{-x+1}\mathrm{d}x = \)

\(-x \cdot e^{-x+1} - e^{-x+1} = e^{-x+1} \cdot (-x-1)\)

\(\int x \cdot e^{2x} \mathrm{d}x \)

\(setze: f(x) = x \ ; g'(x) = e^{2x}\)

\( f'(x) = 1 \ ; g(x) = \frac{1}{2} e^{2x}\)

\(\int x \cdot e^{2x} \mathrm{d}x = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} \cdot e^{2x} \mathrm{d}x = \)

\(x \cdot \frac{1}{2} e^{2x} - \frac{1}{2} \cdot \frac{1}{2} \cdot e^{2x} = e^{2x} \cdot (\frac{x}{2}-\frac{1}{4})\)

\(\int2 x \cdot e^{2x+1} \mathrm{d}x \)

\(setze: f(x) =2 x \ ; g'(x) = e^{2x+1}\)

\( f'(x) = 2 \ ; g(x) = \frac{1}{2} e^{2x+1}\)

\(\int2 x \cdot e^{2x+1} \mathrm{d}x = 2x \cdot \frac{1}{2} e^{2x+1} - \int 2 \cdot \frac{1}{2} \cdot e^{2x+1} \mathrm{d}x = \)

\(x e^{2x+1} - \frac{1}{2} \cdot e^{2x+1} = \)

\(e^{2x+1} \cdot(x-\frac{1}{2})\)

\(\int \frac{1}{x} \cdot \ln(x) \mathrm{d}x \)

\(setze: f(x) =\ln(x) \ ; g'(x) = \frac{1}{x}\)

\( f'(x) = \frac{1}{x} \ ; g(x) = \ln(x)\)

\(\int \frac{1}{x} \cdot \ln(x) \mathrm{d}x = \ln(x) \cdot \ln(x) - \int \frac{1}{x} \cdot \ln(x) \mathrm{d}x\)

umformen als Gleichung

\(\int \frac{1}{x} \cdot \ln(x) \mathrm{d}x + \int \frac{1}{x} \cdot \ln(x) \mathrm{d}x = \ln(x) \cdot \ln(x) \)

\(2 \cdot \int \frac{1}{x} \cdot \ln(x) \mathrm{d}x = ( \ln(x))^2\)

\( \int \frac{1}{x} \cdot \ln(x) \mathrm{d}x =\frac{ ( \ln(x))^2}{2}\)

 \(\int x \cdot \ln (x) \ \mathrm{d}x\)

\(f'(x) = x \ , \ g(x) = \ln(x) \)

\(f(x) = \frac{x^2}{2} \ , \ g'(x) = \frac{1}{x}\)

\(\int x \cdot \ln (x) \ \mathrm{d}x = \frac{x^2}{2} \cdot \ln(x) - \int \frac{1}{x} \cdot \frac{x^2}{2}\mathrm{d}x = \\ \frac{x^2}{2} \cdot \ln(x) - \int \frac{x}{2}\mathrm{d}x = \\ \)

\( \frac{x^2}{2} \cdot \ln(x) - \frac{1}{2} \cdot \frac{x^2}{2}= \\\)

\(\frac{x^2}{2} \cdot (\ln(x) - \frac{1}{2} ) \)

\(\int \limits_{1}^{4} x \cdot \ln (x) \ \mathrm{d}x = [ \frac{x^2}{2} \cdot (\ln(x) - \frac{1}{2} ) ]^4_1 = \)

\(\frac{16}{2} \cdot (\ln(4) - \frac{1}{2} ) - \frac{1}{2} \cdot (\ln(1) - \frac{1}{2} ) = 8 \cdot (\ln(4) - \frac{1}{2} ) - \frac{1}{2} \cdot (\ln(1) - \frac{1}{2} ) =\)

\(= \ln(4^8) - 4 + \frac{1}{4} = \ln(4^8) - \frac{15}{4}\)

\(\int x \cdot e^{-x} \mathrm{d}x \)

\(setze: f(x) = x \ ; g'(x) = e^{-x}\)

\( f'(x) = 1 \ ; g(x) = - e^{-x}\)

\(\int x \cdot e^{-x} \mathrm{d}x = - x \cdot e^{-x} - \int 1 \cdot (-1) \cdot e^{-x} \mathrm{d}x = \)

\(-x \cdot e^{-x} - e^{-x} = e^{-x} \cdot (-x-1)\)

\(\int\limits_{0}^{\infty} x \cdot e^{-x} \mathrm{d}x =\lim_{n \to \infty} \int\limits_{0}^{n} x \cdot e^{-x} \mathrm{d}x =\)

\(\lim_{n \to \infty} [ e^{-n} \cdot (-n-1) - e^{0} \cdot (-1)] = \)

\(\lim_{n \to \infty} [ \frac{-n-1}{e^{n}}+1] = 0+1 = 1\)

\(\int x \cdot \cos(x^2) \ \mathrm{d}x\)

\(\text{setze } u=x^2\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x^2)\)

\(\mathrm{d}x=\frac {1}{2x} \mathrm{d}u\)

\(\int x \cdot \cos(x^2) \ \mathrm{d}x = \int x \cdot \cos(u) \cdot \frac1{2x} \ \mathrm{d}u=\)

\(\frac12 \cdot \int \cos(u) \ \mathrm{d}u = \frac12 \cdot \sin(u) \ = \)

\(\frac12 \cdot \sin(x^2)\)

\(\int x \cdot \sin(x^2) \ \mathrm{d}x\)

\(\text{setze } u=x^2\)

\(\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x^2)\)

\(\mathrm{d}x=\frac {1}{2x} \mathrm{d}u\)

\(\int x \cdot \sin(x^2) \ \mathrm{d}x = \int x \cdot \sin(u) \cdot \frac1{2x} \ \mathrm{d}u=\)

\(\frac12 \cdot \int \sin(u) \ \mathrm{d}u = - \frac12 \cdot \cos(u) \ = \)

\(-\frac12 \cdot \cos(x^2)\)

\(\int e^{-ax} \cdot \sin(x) \ \mathrm{d}x\)

\(setze: f(x) =e^{-ax} \ ; g'(x) = \sin(x)\)

\( f'(x) = -a \cdot e^{-ax} \ ; g(x) = - \cos(x)\)

\(\int e^{-ax} \cdot \sin(x) \ \mathrm{d}x = -\cos(x) \cdot e^{-ax} -a \int e^{-ax} \cdot \cos(x) \ \mathrm{d}x =\)

\(\text{nochmalige partielle Integration, setze} f(x) =e^{-ax} \ ; g'(x) = \cos(x)\)

\( f'(x) = -a \cdot e^{-ax} \ ; g(x) = \sin(x)\)

\(-\cos(x) \cdot e^{-ax} - a [ e^{-ax} \cdot \sin(x) + a \cdot \int e^{-ax} \cdot \sin(x) \ \mathrm{d}x] = \)

\(e^{-ax} \cdot ( - \cos(x) - a \sin(x) ) - a^2 \cdot \int e^{-ax} \cdot \sin(x) \ \mathrm{d}x \)

\(\text{d.h.} \int e^{-ax} \cdot \sin(x) \ \mathrm{d}x = e^{-ax} \cdot ( - \cos(x) -a \sin(x) ) - a^2 \cdot \int e^{-ax} \cdot \sin(x) \ \mathrm{d}x \)

\(\text{umformen als Gleichung ergibt}\)

\((1+a^2) \cdot \int e^{-ax} \cdot \sin(x) \ \mathrm{d}x = e^{-ax} \cdot ( - \cos(x) - a \sin(x) )\)

\(\int e^{-ax} \cdot \sin(x) \ \mathrm{d}x = \frac{e^{-ax} \cdot ( - \cos(x) - a \sin(x) )}{1+a^2 }\)

\(\int \sin(x) \cdot \cos(x) \ \mathrm{d}x\)

\(setze: f(x) =\sin(x) \ ; g'(x) = \cos(x)\)

\( f'(x) = \cos(x) \ ; g(x) = \sin(x) \)

\(\int \sin(x) \cdot \cos(x) \ \mathrm{d}x = \sin(x) \cdot \sin(x) - \int \sin(x) \cdot \cos(x) \ \mathrm{d}x\)

\(\text{umformen als Gleichung ergibt}\)

\(2 \cdot \int \sin(x) \cdot \cos(x) \ \mathrm{d}x = (\sin(x))^2\)

\(\int \sin(x) \cdot \cos(x) \ \mathrm{d}x = \frac{(\sin(x))^2}2\)

\(\text{wird im ersten Schritt alternativ} \ f(x) =\cos(x) \ ; g'(x) = \sin(x) \ \text{gesetzt, so ist das Ergebnis}\)

\(-\frac{(\cos(x))^2}2\)

\(\int \sin(x) \cdot \cos(x) \ \mathrm{d}x\)

\(setze: f(x) =\sin(x) \ ; g'(x) = \cos(x)\)

\( f'(x) = \cos(x) \ ; g(x) = \sin(x) \)

\(\int \sin(x) \cdot \cos(x) \ \mathrm{d}x = \sin(x) \cdot \sin(x) - \int \sin(x) \cdot \cos(x) \ \mathrm{d}x\)

\(\text{umformen als Gleichung ergibt}\)

\(2 \cdot \int \sin(x) \cdot \cos(x) \ \mathrm{d}x = (\sin(x))^2\)

\(\int \sin(x) \cdot \cos(x) \ \mathrm{d}x = \frac12 \cdot (\sin(x))^2\)

\(\int \limits_{0}^{\frac\pi2} \sin(x) \cdot \cos(x) \ \mathrm{d}x = [\frac12 (\sin(x))^2]_{0}^{\frac\pi2} = \)

\( \frac12 (\sin(\frac\pi2))^2-\frac12 (\sin(0))^2 = \frac12\)

\(\int (sin(x))^2 \mathrm{d}x = \int sin(x) \cdot sin(x) \mathrm{d}x \)

\(setze: f(x) = \sin(x) \ ; g'(x) = \sin(x)\)

\( f'(x) = \cos(x) \ ; g(x) = - \cos(x) \)

\(\int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (cos(x))^2\mathrm{d}x \)

\(\int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (1-(\sin(x))^2) \mathrm{d}x \)

Umformen als Gleichung

\(\int (sin(x))^2 \mathrm{d}x + \int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int 1 \mathrm{d}x \)

\(2 \cdot \int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + x \)

\(\int (sin(x))^2 \mathrm{d}x =\frac12 \cdot ( - \sin(x) \cdot \cos(x) + x) \)

\(\int (sin(x))^2 \mathrm{d}x = \int sin(x) \cdot sin(x) \mathrm{d}x \)

\(setze: f(x) = \sin(x) \ ; g'(x) = \sin(x)\)

\( f'(x) = \cos(x) \ ; g(x) = - \cos(x) \)

\(\int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (cos(x))^2\mathrm{d}x \)

\(\int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (1-(\sin(x))^2) \mathrm{d}x \)

Umformen als Gleichung

\(\int (sin(x))^2 \mathrm{d}x + \int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int 1 \mathrm{d}x \)

\(2 \cdot \int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + x \)

\(\int (sin(x))^2 \mathrm{d}x =\frac12 \cdot ( - \sin(x) \cdot \cos(x) + x) \)

\(\int \limits_0^{\frac{\pi}2} (sin(x))^2 \mathrm{d}x = [\frac12 \cdot ( - \sin(x) \cdot \cos(x) + x)]_0^{\frac{\pi}2} = \frac{\pi}4 \)

\(\int (sin(x))^2 \mathrm{d}x = \int sin(x) \cdot sin(x) \mathrm{d}x \)

\(setze: f(x) = \sin(x) \ ; g'(x) = \sin(x)\)

\( f'(x) = \cos(x) \ ; g(x) = - \cos(x) \)

\(\int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (cos(x))^2\mathrm{d}x \)

\(\int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (1-(\sin(x))^2) \mathrm{d}x \)

Umformen als Gleichung

\(\int (sin(x))^2 \mathrm{d}x + \int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int 1 \mathrm{d}x \)

\(2 \cdot \int (sin(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + x \)

\(\int (sin(x))^2 \mathrm{d}x =\frac12 \cdot ( - \sin(x) \cdot \cos(x) + x) \)

\(\int \limits_0^{\pi} (sin(x))^2 \mathrm{d}x = [\frac12 \cdot ( - \sin(x) \cdot \cos(x) + x)]_0^{\pi} = \frac{\pi}2 \)

\(\int (\cos(x))^2 \mathrm{d}x = \int \cos(x) \cdot \cos(x) \mathrm{d}x \)

\(setze: f(x) = \cos(x) \ ; g'(x) = \cos(x)\)

\( f'(x) = -\sin(x) \ ; g(x) =\sin(x)\)

\(\int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + \int (\sin(x))^2\mathrm{d}x \)

\(\int (\cos(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (1-(\cos(x))^2) \mathrm{d}x \)

Umformen als Gleichung

\(\int (\cos(x))^2 \mathrm{d}x + \int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + \int 1 \mathrm{d}x \)

\(2 \cdot \int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + x \)

\(\int (\cos(x))^2 \mathrm{d}x =\frac12 \cdot ( \sin(x) \cdot \cos(x) + x) \)

\(\int (\cos(x))^2 \mathrm{d}x = \int \cos(x) \cdot \cos(x) \mathrm{d}x \)

\(setze: f(x) = \cos(x) \ ; g'(x) = \cos(x)\)

\( f'(x) = -\sin(x) \ ; g(x) =\sin(x)\)

\(\int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + \int (\sin(x))^2\mathrm{d}x \)

\(\int (\cos(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (1-(\cos(x))^2) \mathrm{d}x \)

Umformen als Gleichung

\(\int (\cos(x))^2 \mathrm{d}x + \int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + \int 1 \mathrm{d}x \)

\(2 \cdot \int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + x \)

\(\int (\cos(x))^2 \mathrm{d}x =\frac12 \cdot ( \sin(x) \cdot \cos(x) + x) \)

\(\int \limits_0^{\frac{\pi}2} (\cos(x))^2 \mathrm{d}x =[\frac12 \cdot ( \sin(x) \cdot \cos(x) + x)]_0^{\frac{\pi}2} = \frac{\pi}4\)

\(\int (\cos(x))^2 \mathrm{d}x = \int \cos(x) \cdot \cos(x) \mathrm{d}x \)

\(setze: f(x) = \cos(x) \ ; g'(x) = \cos(x)\)

\( f'(x) = -\sin(x) \ ; g(x) =\sin(x)\)

\(\int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + \int (\sin(x))^2\mathrm{d}x \)

\(\int (\cos(x))^2 \mathrm{d}x = - \sin(x) \cdot \cos(x) + \int (1-(\cos(x))^2) \mathrm{d}x \)

Umformen als Gleichung

\(\int (\cos(x))^2 \mathrm{d}x + \int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + \int 1 \mathrm{d}x \)

\(2 \cdot \int (\cos(x))^2 \mathrm{d}x = \sin(x) \cdot \cos(x) + x \)

\(\int (\cos(x))^2 \mathrm{d}x =\frac12 \cdot ( \sin(x) \cdot \cos(x) + x) \)

\(\int \limits_0^{\pi} (\cos(x))^2 \mathrm{d}x =[\frac12 \cdot ( \sin(x) \cdot \cos(x) + x)]_0^{\pi} = \frac{\pi}2\)

\(\int x \cdot \cos(x) \ \mathrm{d}x\)

\(f(x) = x \ , \ g'(x) = \cos(x) \)

\(f(x) = 1 \ , \ g'(x) = \sin(x)\)

\(\int x \cdot \cos(x) \ \mathrm{d}x = x \sin(x) - \int \sin(x) \ \mathrm{d}x = x \sin(x) + \cos(x)\)

\(\int x \cdot \sin(x) \ \mathrm{d}x\)

\(f(x) = x \ , \ g'(x) = \sin(x) \)

\(f(x) = 1 \ , \ g'(x) = -\cos(x)\)

\(\int x \cdot \sin(x) \ \mathrm{d}x = - x \cos(x) + \int \cos(x) \ \mathrm{d}x = - x \cos(x) + \sin(x)\)

\(\int x^2 \cdot cos(x) \mathrm{d}x \)

\(setze: f(x) = x^2 \ ; g'(x) = \cos(x)\)

\( f'(x) = 2x \ ; g(x) =\sin(x)\)

\(\int x^2 \cdot cos(x) \mathrm{d}x = x^2 \sin(x) - \int 2x\sin(x) \mathrm{d}x \)

nochmalige partielle Integration

\(setze: f(x) = 2x \ ; g'(x) = \sin(x)\)

\( f'(x) = 2 \ ; g(x) =-\cos(x)\)

\( x^2 \sin(x) - \int 2x\sin(x) \mathrm{d}x = x^2 \sin(x) + 2x\cos(x) - 2 \sin(x)\)

\(\int x^2 \cdot \sin(x) \mathrm{d}x \)

\(setze: f(x) = x^2 \ ; g'(x) = \sin(x)\)

\( f'(x) = 2x \ ; g(x) =-\cos(x)\)

\(\int x^2 \cdot \sin(x) \mathrm{d}x = -x^2 \cos(x) + \int 2x \cos(x) \mathrm{d}x = \)

nochmalige partielle Integration

\(setze: f(x) = 2x \ ; g'(x) = \cos(x)\)

\( f'(x) = 2 \ ; g(x) =\sin(x)\)

\( -x^2 \cos(x) + \int 2x \cos(x) \mathrm{d}x = -x^2 \cos(x) + 2x\sin(x) - \int 2 \sin(x) \mathrm{d}x = -x^2 \cos(x) + 2x\sin(x) + 2 \cos(x) \)

\(\int x^2 \cdot \sin(x) \mathrm{d}x \)

\(setze: f(x) = x^2 \ ; g'(x) = \sin(x)\)

\( f'(x) = 2x \ ; g(x) =-\cos(x)\)

\(\int x^2 \cdot \sin(x) \mathrm{d}x = -x^2 \cos(x) + \int 2x \cos(x) \mathrm{d}x = \)

nochmalige partielle Integration

\(setze: f(x) = 2x \ ; g'(x) = \cos(x)\)

\( f'(x) = 2 \ ; g(x) =\sin(x)\)

\( -x^2 \cos(x) + \int 2x \cos(x) \mathrm{d}x = -x^2 \cos(x) + 2x\sin(x) - \int 2 \sin(x) \mathrm{d}x = \)

\( -x^2 \cos(x) + 2x\sin(x) + 2 \cos(x) \)

\(\int\limits_{0}^\pi x^2 \cdot \sin(x) \mathrm{d}x = [ -x^2 \cos(x) + 2x\sin(x) + 2 \cos(x) ]_{0}^\pi = \)

\( -\pi^2 \cos(\pi) + 2\pi\sin(\pi) + 2 \cos(\pi) - 2 \cos(0) = \pi^2 -4\)