From Förberedande kurs i matematik 2

Revision as of 14:47, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.3:1c moved to Solution 2.3:1c: Robot: moved page) ← Previous diff		Current revision (08:22, 29 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The integrand consists of two factors, so integration by parts is a plausible method. The most obvious thing to do is to choose <math>x^2</math> as the factor that we will differentiate and <math>\cos x</math> as the factor that we will integrate. Admittedly, the <math>x^2</math>-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,
-	<center> [[Image:2_3_1c-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:2_3_1c-2(2).gif]] </center>	+	We can attack the integral on the right-hand side in the same way. Let <math>2x</math> be the factor that we differentiate and <math>\sin x</math> the factor that we integrate. This time, we have only one factor left,
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\begin{align}
		+	\int 2x\cdot \sin x\,dx
		+	&= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt]
		+	&= -2x\cos x + 2\int \cos x\,dx\\[5pt]
		+	&= -2x\cos x + 2\sin x + C\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	All in all, we obtain
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int x^2\cos x\,dx
		+	&= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt]
		+	&= x^2\sin x + 2x\cos x - 2\sin x + C\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.

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Current revision

The integrand consists of two factors, so integration by parts is a plausible method. The most obvious thing to do is to choose \displaystyle x^2 as the factor that we will differentiate and \displaystyle \cos x as the factor that we will integrate. Admittedly, the \displaystyle x^2-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,

\displaystyle \int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}

We can attack the integral on the right-hand side in the same way. Let \displaystyle 2x be the factor that we differentiate and \displaystyle \sin x the factor that we integrate. This time, we have only one factor left,

\displaystyle \begin{align} \int 2x\cdot \sin x\,dx &= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt] &= -2x\cos x + 2\int \cos x\,dx\\[5pt] &= -2x\cos x + 2\sin x + C\,\textrm{.} \end{align}

All in all, we obtain

\displaystyle \begin{align} \int x^2\cos x\,dx &= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt] &= x^2\sin x + 2x\cos x - 2\sin x + C\,\textrm{.} \end{align}

For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.