Question

..... next set of homeworks :)

Answer 1

trig subs .... booyaaa!!!

Answer 2

better you than i
or is it "better you than me"?

Answer 3

not a trig sub... a reduction or something

Answer 4

\[\Large
\begin{align}

2)&\ \int sin^6(x)cos^3(x)\ dx\\\\
&=\int sin^6(x)cos^2(x)cos(x)\ dx\\\\
&=\int sin^6(x)(1-sin^2(x))cos(x)\ dx\\\\
&=\int (sin^6(x)-sin^8(x))cos(x)\ dx\\\\
&=\int sin^6(x)cos(x)-sin^8(x)cos(x)\ dx\\\\
&=\frac{sin^7(x)}{7}-\frac{sin^9(x)}{9}+C\\\\

\end{align}\]

Answer 5

\[\Large
\begin{align}

4)&\ \int_{0}^{\pi/2} cos^5(x)\ dx\\\\
&=\frac{1}{5}cos^{4}(x)sin(x)+\frac{4}{5}\int cos^{3}(x)dx\\\\
&\left(\frac{1}{3}cos^{2}(x)sin(x)+\frac{2}{3}\int cos(x)dx\right)\\\\
&\left.(sin(x))\ \right|_{0}^{\pi/2}\\\\
\end{align}\]
ugh .... now i got to get a number out of that?

\[\Large
\begin{align}
&\frac{1}{5}cos^{4}(\pi/2)sin(\pi/2)+\frac{4}{5}\\\\
&\left(\frac{1}{3}cos^{2}(\pi/2)sin(\pi/2)+\frac{2}{3}\ \right)\\\\
&(sin(\pi/2))-\\\\
&\frac{1}{5}cos^{4}(0)sin(0)+\frac{4}{5}\\\\
&\left(\frac{1}{3}cos^{2}(0)sin(0)+\frac{2}{3}\ \right)\\\\
&(sin(0))\\\\

&=0
\end{align}\]
gonna half to chk that with the wolf

Answer 6

\[\Large \frac{8}{15}\]

Answer 7

\[\Large
\begin{align}

6)&\ \int \frac{sin^3(\sqrt{x})}{\sqrt{x}}\ dx\\\\
&u=\sqrt{x};\ du=\frac{dx}{2\sqrt{x}};\ 2du=\frac{dx}{\sqrt{x}}\\\\
&=\int 2\ sin^3(u)\ du\\\\
&=2*-\frac{1}{3}(2+sin^2(u))cos(u)+C\\\\
&=-\frac{2}{3}(2+sin^2(\sqrt{x}))cos(\sqrt{x})+C\\\\

\end{align}\]

Answer 8

\[\Large
\begin{align}

8)&\ \int_{0}^{\pi/2} sin^2(2t)\ dt\\\\
&=\int_{0}^{\pi/2} \frac{1}{2}-\frac{cos(4t)}{2}\ dt\\\\
&= \left.\frac{t}{2}-\frac{sin(4t)}{8}\ \right|_{0}^{\pi/2}\\\\
&= \frac{\pi}{4}-\frac{sin(2\pi)}{8}\\\\
&-\frac{0}{2}+\frac{sin(4(0))}{8}\\\\
&= \frac{\pi}{4}\\\\



\end{align}\]

Answer 9

\[\Large
\begin{align}

10)&\ \int_{0}^{\pi} cos^6(t)\ dt\\\\
&= \frac{1}{6}cos^5(t)sin(t)+\frac{5}{6}\\\\
& \left(\frac{1}{4}cos^3(t)sin(t)+\frac{3}{4}\ \ \right)\\\\
& \left(\frac{1}{2}cos(t)sin(t)+\frac{t}{2}\right)\\\\
&\left.\frac{}{}\right|_{0}^{\pi}\\\\

&= \frac{1}{6}cos^5(\pi)sin(\pi)+\frac{5}{6}\\\\
& \left(\frac{1}{4}cos^3(\pi)sin(\pi)+\frac{3}{4}\ \ \right)\\\\
& \left(\frac{1}{2}cos(\pi)sin(\pi)+\frac{\pi}{2}\right)\\\\

&- \frac{1}{6}cos^5(0)sin(0)+\frac{5}{6}\\\\
& \left(\frac{1}{4}cos^3(0)sin(0)+\frac{3}{4}\ \ \right)\\\\
& \left(\frac{1}{2}cos(0)sin(0)+\frac{0}{2}\right)\\\\

\end{align}\]

Answer 10

\[\Large \frac{5\pi}{16}\]

Answer 11

13
14
16
19
20
21
24
26
27
36

left to finish

Answer 12

\[\Large
\begin{align}

13)&\ \int_{0}^{\pi/2} sin^2(x)cos^2(x)\ dx\\\\
&= \int_{0}^{\pi/2} sin^2(x)(1-sin^2(x))\ dx\\\\
&= \int_{0}^{\pi/2} sin^2(x)-sin^4(x)\ dx\\\\
&= [shortcut]\\\\
&= \frac{1}{2}\frac{\pi}{2}-\frac{3}{4}\frac{1}{2}\frac{\pi}{2}\\\\
&= \frac{\pi}{4}-\frac{3\pi}{16}=\frac{\pi}{16}\\\\

\end{align}\]

Answer 13

\[\Large
\begin{align}

14)&\ \int_{0}^{\pi} sin^2(x)cos^4(x)\ dx\\\\
&= \int_{0}^{\pi} (1-cos^2(x))cos^4(x)\ dx\\\\
&= \int_{0}^{\pi} cos^4(x)-cos^6(x)\ dx\\\\
&= [shortcut]\\\\
&= \frac{3}{4}\frac{1}{2}\frac{\pi}{1}-\frac{5}{6}\frac{3}{4}\frac{1}{2}\frac{\pi}{1}\\\\
&= \frac{3\pi}{8}-\frac{5\pi}{16}=\frac{\pi}{16}\\\\

\end{align}\]

Answer 14

\[\Large
\begin{align}

16)&\ \int cos^5(u)\ du;\ u=sin(u);\ du=cos(u)\\\\
&= \frac{cos^4(u)sin(u)}{5}+\frac{4}{5}\left(\ \frac{t}{b}\ \right)\\\\
&\left(\frac{cos^2(u)sin(u)}{3}+\frac{2}{3}sin(u)\right)\\\\
&= \frac{cos^4(sin(t))sin(sin(t))}{5}+\frac{4}{5}\left(\ \frac{t}{b}\ \right)\\\\
&\left(\frac{cos^2(sin(t))sin(sin(t))}{3}+\frac{2}{3}sin(T)\right)\\\\
&+\frac{2}{3}sin(sin(t))\ +C\\\\
\end{align}\]

Answer 15

\[\Large
\begin{align}

19)&\ \int \frac{cos(x)+sin(2x)}{sin(x)}\ dx\\\\
&=\int \frac{cos(x)}{sin(x)}+\frac{sin(2x)}{sin(x)}\ dx\\\\
&=\int \frac{cos(x)}{sin(x)}+\frac{2sin(x)cos(x)}{sin(x)}\ dx\\\\
&=\int \frac{cos(x)}{sin(x)}+2cos(x)\ dx\\\\
&=ln(sin(x))+2sin(x)+C\\\\
\end{align}\]

\[\Large
\begin{align}

20)&\ \int cos^2(x)+sin(2x)\ dx\\\\
&= \int \frac{1-cos(2x)}{2}\\\\
&+2sin(x)cos(x)\ dx\\\\
&= \frac{x}{2}-\frac{cos(2x)}{4}\\\\
&+sin^2(x)+C\\\\


\end{align}\]