Finish assignment 5

1 files changed, 26 insertions, 1 deletions

diff --git a/assignment5.tex b/assignment5.tex
index cf7d376..4447a5e 100644
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@@ -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}