From 100c0361e4b4bb8becbfe7b90d6e119599b6d070 Mon Sep 17 00:00:00 2001 From: Camil Staps Date: Wed, 7 Oct 2015 10:07:14 +0200 Subject: Finish assignment 5 --- assignment5.tex | 27 ++++++++++++++++++++++++++- 1 file changed, 26 insertions(+), 1 deletion(-) diff --git a/assignment5.tex b/assignment5.tex index cf7d376..4447a5e 100644 --- a/assignment5.tex +++ b/assignment5.tex @@ -41,7 +41,32 @@ But then also: $2\cdot\int\cos^2x\diff x = \sin x\cdot\cos x + x + c$, so $\int\cos^2x\diff x = \frac12\left(\sin x\cdot\cos x + x\right) + c'$. \item $\int\frac1{\sqrt{1-4x^2}}\diff x = \int\frac1{\sqrt{1-u^2}}\cdot\frac{\diff\frac12u}{\diff u}\diff u = \int\frac1{\sqrt{1-u^2}}\cdot\frac12\diff u = \frac12\arcsin(2x) + c$. - \item %todo + \item Using integration by parts twice, we find: + + \begin{equation} + \int e^{3x}\sin x\diff x = \tfrac13e^{3x}\sin x - \tfrac13\int e^{3x}\cos x\diff x. + \label{eq:1ea} + \end{equation} + and + \begin{equation*} + \int e^{3x}\sin x\diff x = -e^{3x}\cos x + 3\int e^{3x}\cos x\diff x. + \end{equation*} + + This then means: + + \begin{align*} + \tfrac13e^{3x}\sin x - \tfrac13\int e^{3x}\cos x\diff x &= -e^{3x}\cos x + 3\int e^{3x}\cos x\diff x \\ + e^{3x}\left(\tfrac13\sin x+\cos x\right) &= 3\tfrac13\int e^{3x}\cos x\diff x \\ + \int e^{3x}\cos x\diff x &= \tfrac1{10}e^{3x}(\sin x+3\cos x). + \end{align*} + + Substituting that into equation \ref{eq:1ea} gives: + + \begin{align*} + \int e^{3x}\sin x\diff x &= \tfrac13e^{3x}\sin x - \tfrac1{30}e^{3x}(\sin x+3\cos x) + c \\ + &= -\tfrac1{30}e^{3x}(3\cos x-9\sin x) + c \\ + &= -\tfrac1{10}e^x(\cos x-3\sin x) + c. + \end{align*} \end{enumerate} \item \begin{enumerate} -- cgit v1.2.3