Question is self explanatory; integration of the above (I tried a substitution of $\sin(x)=u$ after getting the integral of $\sin(101x)\cdot \sin^{99}x$.
I have a complicated integral to solve. I tried to split ($101 x$) and proceed but I am getting a pretty nasty answer while evaluating using parts. are there any simpler methods to evaluate this integral?  $$ \int\!\sin (101x)\cdot\sin^{99}(x)\, dx $$
I installed Ubuntu 17.10. Now I am having trouble with gksu:  $ gksu -dg synaptic No ask_pass set, using default! xauth: /tmp/libgksu-HgUjgQ/.Xauthority STARTUP_ID: gksu/synaptic/8760-0-alex-XPS-15-9530_TIME4974977 cmd[0]: /usr/bin/sudo cmd[1]: -H cmd[2]: -S cmd[3]: -p cmd[4]: GNOME_SUDO_PASS cmd[5]: -u cmd[6]: root cmd[7]: -- cmd[8]: synaptic buffer: -GNOME_SUDO_PASS- brute force GNOME_SUDO_PASS ended... Yeah, we're in... Unable to init server: Could not connect: Connection refused (synaptic:8767): Gtk-WARNING **: cannot open display: :1 xauth: /tmp/libgksu-HgUjgQ/.Xauthority xauth_env: (null) dir: /tmp/libgksu-HgUjgQ   If I don't use -g, the password dialog is disabled. So looks like a problem with creating a tty for root.  Any advice? How to set a command block with a command inside using the setblock command Does someone know how to set a command block with a command inside using the setblock command?  I have created this code /setblock &lt;x&gt; &lt;y&gt; &lt;z&gt; command_block {TileEntityData:{Command:"testfor @p[_=1]"}} but it doesn't works.  Do someone know how to do that? Use property to show that if $f_n \rightarrow f$ in measure on finite measure set $E$ then $f_n^2 \rightarrow f^2$ in measure on $E$. Consider the following problem:     Let $E$ have finite measure, $\{f_n\} \rightarrow f$ in measure on $E$ and $g$ be a measurable function on $E$ that is finite almost everywhere. Use the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ to prove that $\{f_n^2 \} \rightarrow f^2$ in measure on $E$. Infer from this that if $\{g_n\} \rightarrow g$ in measure on $E$ then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure on $E$.   I'm struggling understand how I can apply the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ in measure to prove that $\{f_n^2 \} \rightarrow f^2$ in measure. I've been able to come up with a proof of the fact that if $\{g_n\} \rightarrow g$ in measure then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure, and this clearly implies the claim that $\{f_n^2 \} \rightarrow f^2$ in measure. Unfortunately my proof doesn't rely on the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ so it doesn't answer the problem since it explicitly asks us to prove it a certain way.  So far my best attempt to utilize the fact is as follows. Let $\eta &gt; 0$ then by triangle inequality and monotonicity of measure  \begin{align*}     \lim_{n\rightarrow \infty}\mu\left\{ \left|f_n^2 - f^2\right| &gt; \eta \right\} &amp;\leq \lim_{n\rightarrow \infty}\mu \left\{\left|f_n - f\right|\cdot \left|f_n\right| + \left|f_n - f\right| \cdot \left|f\right| &gt; \eta \right\} \\     &amp;\leq \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right|\cdot \left|f_n\right| &gt; \eta/2 \right\} + \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| &gt; \eta/2 \right\} \end{align*}  By the fact that $\{f_n \cdot f\} \rightarrow f \cdot f$ in measure \begin{equation*} \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| &gt; \eta/2 \right\} = \mu\left\{\left|f_n \cdot f - f\cdot f\right| &gt; \eta/2 \right\} = 0. \end{equation*} I don't see how to deal with the other limit without first proving the most general statement $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure.   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