Solution 2.2:3a
From Förberedande kurs i matematik 2
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| - | {{ | + | The secret behind a successful substitution is to be able to recognize the integral as an expression of the type |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\int \left( \begin{matrix} |
| + | \text{an expression}\\ | ||
| + | \text{in u} | ||
| + | \end{matrix}\right)\cdot u'\,dx\,,</math>}} | ||
| + | |||
| + | where <math>u=u(x)</math> is the actual substitution. In the integral | ||
| + | |||
| + | {{Displayed math||<math>\int 2x\sin x^2\,dx</math>}} | ||
| + | |||
| + | we see that the expression <math>x^2</math> is the argument for the sine function, as the same time as its derivative <math>\bigl(x^2\bigr)'=2x</math> stands as a factor in front of sine. Therefore, if we set <math>u=x^2</math>, the integral, the integral will be of the form | ||
| + | |||
| + | {{Displayed math||<math>\int u'\sin u\,dx\,\textrm{.}</math>}} | ||
| + | |||
| + | Thus, we can use <math>u=x^2</math> for the substitution, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \int 2x\sin x^2\,dx | ||
| + | &=\left\{\begin{align} | ||
| + | u &= x^2\\[5pt] | ||
| + | du &= 2x\,dx | ||
| + | \end{align}\right\} = \int{\sin u\,du}\\[5pt] | ||
| + | &= -\cos u+C = -\cos x^2 + C\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
The secret behind a successful substitution is to be able to recognize the integral as an expression of the type
| \displaystyle \int \left( \begin{matrix}
\text{an expression}\\ \text{in u} \end{matrix}\right)\cdot u'\,dx\,, |
where \displaystyle u=u(x) is the actual substitution. In the integral
| \displaystyle \int 2x\sin x^2\,dx |
we see that the expression \displaystyle x^2 is the argument for the sine function, as the same time as its derivative \displaystyle \bigl(x^2\bigr)'=2x stands as a factor in front of sine. Therefore, if we set \displaystyle u=x^2, the integral, the integral will be of the form
| \displaystyle \int u'\sin u\,dx\,\textrm{.} |
Thus, we can use \displaystyle u=x^2 for the substitution,
| \displaystyle \begin{align}
\int 2x\sin x^2\,dx &=\left\{\begin{align} u &= x^2\\[5pt] du &= 2x\,dx \end{align}\right\} = \int{\sin u\,du}\\[5pt] &= -\cos u+C = -\cos x^2 + C\,\textrm{.} \end{align} |
