... dont mind me ...

Question

myininaya · Answer

amistre are you posting all of your homework lol

amistre64 · Answer

$$\Large
\begin{align}

21)&\ \int sec^2(x)tan(x) dx\\
&=\ \int (tan^2(x)+1)tan(x) dx\\
&=\ \int tan^3(x)+tan(x) dx\\
&=\ \int tan^3(x)dx-ln|cos(x)|\\
&=\  \frac{1}{2}tan^2(x)+ln|cos(x)|-ln|cos(x)|+C\\
&=\  \frac{1}{2}tan^2(x)+C\\

\end{align}$$

amistre64 · Answer

my handwriting is not that legible so i come here to format it and print it

amistre64 · Answer

$$\Large
\begin{align}

24)&\ \int_{0}^{\pi/3} tan^5(x)sec^4(x)dx\\
&=\int_{0}^{\pi/3} tan^4(x)sec^3(x)\\&
(sec(x)tan(x))dx\\
&=\int_{0}^{\pi/3} (sec^2(x)-1)^2sec^3(x)\\&
(sec(x)tan(x))dx\\
&=\int_{0}^{\pi/3} (sec^4(x)-2sec^2(x)+1)\\&
sec^3(x)(sec(x)tan(x))dx\\
&=\int_{0}^{\pi/3} (sec^7(x)-2sec^5(x)+sec^3(x))\\&
(sec(x)tan(x))dx\\
&=\frac{sec^8(x)}{8}-\frac{sec^6(x)}{3}+\frac{sec^4(x)}{4}\\
&\left.\frac{t}{b}\ \right|_{0}^{\pi/3}\\
&=\frac{sec^8(\pi/3)}{8}-\frac{sec^6(\pi/3)}{3}+\frac{sec^4(\pi/3)}{4}\\
&=-\frac{sec^8(0)}{8}+\frac{sec^6(0)}{3}-\frac{sec^4(0)}{4}\\
&=\frac{2^8}{8}-\frac{2^6}{3}+\frac{2^4}{4}\\
&-\frac{1}{8}+\frac{1}{3}-\frac{1}{4}\\
&=\frac{44}{3}-\frac{1}{24}=\frac{117}{8}\\


\end{align}$$

amistre64 · Answer

$$\Large
\begin{align}

26)&\ \int_{0}^{\pi/4} sec^4(x)tan^4(x) dx\\
&=\int_{0}^{\pi/4} (tan^2(x)+1)^2tan^4(x) dx\\
&=\int_{0}^{\pi/4} (tan^4(x)+2tan^2(x)+1)\\
&tan^4(x) dx\\
&=\int_{0}^{\pi/4} (tan^8(x)+2\int tan^6(x)\\&
+\int tan^4(x))dx\\
&=\frac{1}{7}tan^7(x)+\int tan^6(x)\\
&+\int tan^4(x))dx\\

&=\frac{1}{7}tan^7(x)+\frac{1}{5} tan^5(x)\\
\left.\frac{t}{b}\ \right|_{0}^{\pi/4}\\
&=\frac{1}{7}tan^7(\pi/4)+\frac{1}{5} tan^5(\pi/4)\\
&-\frac{1}{7}tan^7(0)-\frac{1}{5} tan^5(0)\\

&=\frac{1}{7}+\frac{1}{5}=\frac{12}{35}\\



\end{align}$$

amistre64 · Answer

$$\Large
\begin{align}

27)&\ \int_{0}^{\pi/3} tan^5(x)sec^4(x) dx\\
&=\int_{0}^{\pi/3} tan^5(x)(tan^2+1)^2 dx\\
&=\int_{0}^{\pi/3} tan^5(x)(tan^4\\
&+2tan^2(x)+1)^2 dx\\
&=\int_{0}^{\pi/3} (tan^9+2tan^7(x)+tan^5(x)) dx\\
&=\frac{1}{8}tan^8(x)+\int_{0}^{\pi/3} tan^7(x)\\
&+\int_{0}^{\pi/3} tan^5(x)) dx\\
&=\frac{1}{8}tan^8(x)+\frac{1}{6}tan^6(x)\\
&\left.\frac{t}{b}\ \right|_{0}^{\pi/3}\\
&=\frac{1}{8}tan^8(\pi/3)+\frac{1}{6}tan^6(\pi/3)\\
&-\frac{1}{8}tan^8(0)-\frac{1}{6}tan^6(0)\\
&=\frac{3^{8/2}}{8}+\frac{3^{6/2}}{6}\\
&=\frac{3^{4}}{8}+\frac{3^{3}}{6}\\
&=\frac{81}{8}+\frac{9}{2}\\
&=\frac{81+36}{8}=\frac{117}{8}\\



\end{align}$$

amistre64 · Answer

ahh one more

amistre64 · Answer

$$\Large
\begin{align}

36)&\ \int \frac{sin(t)}{cos^3(t)}\ dt\\
&=\ \int sin(t)sec^3(t)\ dt\\
&=\ \int tan(t)sec^2(t)\ dt\\
&=\ \frac{tan^2(t)}{2}+C\\

\end{align}$$

amistre64 · Answer

done :)

amistre64 · Answer

i just noticed that the first and last problems were the same lol

amistre64 · Answer

sigh ... gotta get to class, have fun ;)