Solution 2.2:3b
From Förberedande kurs i matematik 2
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| - | {{ | + | If we are to succeed in simplifying the integral with a substitution, we must find an expression <math>u = u(x)</math> so that the integral can be written as |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\int \left(\begin{matrix} |
| + | \text{something}\\ | ||
| + | \text{in u} | ||
| + | \end{matrix}\right)\cdot {u}'\,dx\,\textrm{.}</math>}} | ||
| + | |||
| + | As our integral is written, | ||
| + | |||
| + | {{Displayed math||<math>\int\sin x\cos x\,dx</math>}} | ||
| + | |||
| + | we see that the second factor <math>\cos x</math> is a derivative of the first factor, <math>\sin x</math>. If <math>u=\sin x</math>, the integral can thus be written as | ||
| + | |||
| + | {{Displayed math||<math>\int u\cdot u'\,dx</math>}} | ||
| + | |||
| + | and this makes <math>u=\sin x</math> an appropriate substitution, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \int \sin x\cos x\,dx | ||
| + | &= \left\{ \begin{align} | ||
| + | u &= \sin x\\[5pt] | ||
| + | du &= (\sin x)'\,dx = \cos x\,dx | ||
| + | \end{align} \right\}\\[5pt] | ||
| + | &= \int u\,du\\[5pt] | ||
| + | &= \frac{1}{2}u^{2} + C\\[5pt] | ||
| + | &= \frac{1}{2}\sin^2\!x + C\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
If we are to succeed in simplifying the integral with a substitution, we must find an expression \displaystyle u = u(x) so that the integral can be written as
| \displaystyle \int \left(\begin{matrix}
\text{something}\\ \text{in u} \end{matrix}\right)\cdot {u}'\,dx\,\textrm{.} |
As our integral is written,
| \displaystyle \int\sin x\cos x\,dx |
we see that the second factor \displaystyle \cos x is a derivative of the first factor, \displaystyle \sin x. If \displaystyle u=\sin x, the integral can thus be written as
| \displaystyle \int u\cdot u'\,dx |
and this makes \displaystyle u=\sin x an appropriate substitution,
| \displaystyle \begin{align}
\int \sin x\cos x\,dx &= \left\{ \begin{align} u &= \sin x\\[5pt] du &= (\sin x)'\,dx = \cos x\,dx \end{align} \right\}\\[5pt] &= \int u\,du\\[5pt] &= \frac{1}{2}u^{2} + C\\[5pt] &= \frac{1}{2}\sin^2\!x + C\,\textrm{.} \end{align} |
