just typing up me homeworkage ....

Question

amistre64 · Answer

latex is a pain to type but more legible than my handwriting

amistre64 · Answer

$$\large\begin{array}l
8)\ \int x^2(x^3+5)^9\ dx&=\frac{1}{3}\int3x^2(x^3+5)^9\ dx\\
&=\frac{1}{3(10)}(x^3+5)^{10}+C\\
&=\frac{1}{30}(x^3+5)^{10}+C\
\end{array}$$

$$\large\begin{array}l
12)\ \int \frac{x}{(x^2+1)^2} dx&=\frac12\int{\frac{2x}{(x^2+1)^2}}\ dx\\
&=\frac{1}{2}\frac{-1}{(x^2+1)}+C\\
&=-\frac{1}{2(x^2+1)}+C
\end{array}$$

$$\large\begin{array}l
14)\ \int e^x sin(e^x) dx&=-cos(e^x)+C\\
\end{array}$$

$$\large\begin{array}l
16)\ \int {\frac{x}{x^2+1}}\ dx&=\frac{1}{2}\int \frac{2x}{x^2+1}\ dx\\
&=\frac{1}{2}ln|x^2+1|+C\\
\end{array}$$

$$\large\begin{array}l
22)\ \int {x^{1/2}\ sin(1+x^{3/2})}\ dx\\
=-\frac{2}{3}cos(1+x^{3/2})+C
\end{array}$$

$$\large\begin{array}l
24)\ \int {(1+tan(t))^5sec^2(t)}\ dt\\
=\frac{1}{6}(1+tan(t))^6+C
\end{array}$$

$$\large\begin{array}l
26)\ \int {e^{cos(t)}sin(t)}\ dt&=-\int {e^{cos(t)}-sin(t)}\ dt\\
&=-e^{cos(t)}+C
\end{array}$$


$$\large\begin{array}l
28)\ \int {\frac{tan^{-1}(x)}{1+x^2}}\ dx&=\frac{1}{2}\int {\frac{2(tan^{-1}(x))}{1+x^2}}\ dx\\
&=\frac{1}{2}(tan^{-1}(x))^2+C
\end{array}$$

$$\large\begin{array}l
30)\ \int {\frac{sin(ln(x))}{x}}\ dx&=-\int {\frac{-sin(ln(x))}{x}}\ dx\\
&=-cos(ln(x))+C
\end{array}$$


$$\large\begin{array}l
32)\ \int {\frac{e^x}{e^x+1}}\ dx&=ln(e^x+1)+C
\end{array}$$


$$\large\begin{array}l
34)\ \int {\frac{cos(\pi/x)}{x^2}}\ dx&={-\frac{1}{\pi}\int\frac{-\pi\ cos(\pi/x)}{x^2}}\ dx\\
&=-\frac{sin(\pi/x)}{\pi}+C
\end{array}$$


$$\large\begin{array}l
40)\ \int {sin(t)\ sec^2(t)(cos(t))}\ dt\\
=-\int {-sin(t)\ sec^2(t)(cos(t))}\ dt\\
=-\ tan(cos(t))\ +C\\
\end{array}$$


$$\large\begin{array}l
42)\ \int {\frac{x}{1+x^4}}\ dx&=\int {\frac{x}{1+(x^2)^2}}\ dx\\
&=\frac{1}{2}\int {\frac{2x}{1+(x^2)^2}}\ dx\\
&=\frac{tan^{-1}(x^2)}{2}+C\\


\end{array}$$

amistre64 · Answer

$$\large\begin{array}l
52)\ \int_{0}^{7} {(4+3x)^{1/2}}\ dx&=\frac{2}{9}\int_{0}^{7} {\frac{9}{2}(4+3x)^{1/2}}\ dx\\
&=\left.\frac{2}{9}(4+3x)^{3/2}\right|_{0}^{7}\\
&=\frac{2}{9}(4+3(7))^{3/2}-\frac{2}{9}(4)^{3/2}\\
&=\frac{2(125)}{9}-\frac{2(8)}{9}\\
&=\frac{2}{9}(117)=2(13)\\
&=26\\
\end{array}$$

$$\large\begin{array}l
54)\ \int_{0}^{\sqrt{\pi}} {x\ cos(x^2)}\ dx&=\frac{1}{2}\int_{0}^{\sqrt{\pi}} {2x\ cos(x^2)}dx\\
&=\left.{\frac{sin(x^2)}{2}}\right|_{0}^{\sqrt{\pi}}\\
&=\frac{sin(\pi)}{2}-\frac{sin(0)}{2}\\
&=0\\
\end{array}$$

$$\large\begin{array}l
58)\ \int_{0}^{1} {x\ e^{-x^2}}\ dx&=-\frac{1}{2}\int_{0}^{1} -2{x\ e^{-x^2}}dx\\
&=\left.{-\frac{e^{-x^2}}{2}}\right|_{0}^{1}\\
&=-\frac{e^{-1}}{2}+\frac{e^{0}}{2}\\
&=\frac{1}{2}(1-e^{-1})\\
&=\frac{e-1}{2e}\\
\end{array}$$

amistre64 · Answer

$$\large\begin{array}l
2)\ \int_{0}^{2} {(x+2)^{1/2}-\frac{1}{x+1}}\ dx\\
=\left. \frac{2}{3} (x+2)^{3/2}-ln
|x+1| \right]_{0}^{2}\\
=\frac{2}{3} (2+2)^{3/2}-ln
|2+1| -\frac{2}{3} (0+2)^{3/2}+ln
|0+1|\\
=\frac{16}{3}-ln|3| -\frac{4\sqrt{2}}{3}\\
=\frac{16-4\sqrt{2}}{3}-ln|3|\\
\end{array}$$

$$\large\begin{array}l
9)\ \int_{1}^{2} {x^{-1}-x^{-2}}\ dx&=\left. ln|x|+\frac{1}{x}\right]_{1}^{2}\\
&=ln|2|+\frac{1}{2}-ln|1|-\frac{1}{1}\\
&=ln|2|+\frac{1}{2}-1\\
&=ln|2|-\frac{1}{2}\\
\end{array}$$

$$\large\begin{array}l
13)\ 2\int_{0}^{3} {12-x^2 -x^2+6}\ dx\\
=\left.2(18x-\frac{2}{3}x^3)\ \right|_{0}^{3}\\
=2(18(3)-\frac{2}{3}(3)^3)-0\ \\
=2(54-18)=2(36)\\
=72\\
 \end{array}$$

$$\large\begin{array}l
15)\ 2\int_{0}^{\pi/3} {2sin(x)-tan(x)}\ dx\\
=\left. 2({-2cos(x)}+ln|cos(x)|) \right|_{0}^{\pi/3}\\
= 2({-2cos(\pi/3)}+ln|cos(\pi/3)|\\
+2cos(0)-ln|cos(0)|)\\
= 2(-1+ln|\frac{1}{2}|+2)\\
= 2(1+ln(1)-ln(2))\\
= 2-2\ ln(2)\\
 \end{array}$$

$$\large\begin{array}l
16)\ \int {x^3-x-3x}\ dx;\ [-2,0,2]\\
=\int {x^3-4x};\ [-2,0,2]\ \\
=\left.\frac{x^4}{4}-2x^2\right|_{-2}^{0}\ +\ \left.\frac{x^4}{4}-2x^2\right|_{0}^{2}\\
=0-\frac{(-2)^4}{4}+2(-2)^2\ +\ \frac{(2)^4}{4}-2(2)^2-0\\
=8-4\ +\ 8-4\\
=16\\
 \end{array}$$

amistre64 · Answer

$$\large\begin{array}l
16)\ \int_{-2}^{0} {x^3-4x}\ dx+\int_{0}^{2} {4x-x^3}\ dx\\
=\left.\frac{x^4}{4}-2x^2\right|_{-2}^{0} +\left.2x^2-\frac{x^4}{4}\right|_{0}^{2}\\
=0-\frac{(-2)^4}{4}+2(-2)^2+\frac{(2)^4}{4}-2(2)^2-0\\
=-4+8\ +\ 8-4\\
=16\\
 \end{array}$$

aravindg · Answer

:) am i intruding?

amistre64 · Answer

$$\large\begin{array}l
26)\ 2\int_{0}^{2}{x-x^2+2}\ dx\ =2( \left.\frac{x^2}{2}-\frac{x^3}{3}+2x)\ \right|_{0}^{2}\\
=2(\frac{(2)^2}{2}-\frac{(2)^3}{3}+2(2)-0)\\
=12-\frac{16}{3}\\
=\frac{36-16}{3}\\
=\frac{20}{3}\\
\end{array}$$

$$\large\begin{array}l
27)\ \int_{0}^{1} {x-\frac{x}{4}}\ dx+\int_{1}^{2}{\frac{1}{x}-\frac{x}{4}}\ dx\\
=\left.\frac{3}{8}x^2\right|_{0}^{1}\ +\ \left.{ln(x)-\frac{1}{8}x^2}\right|_{1}^{2} \\
=\frac{3}{8}-0+ln(2)-\frac{1}{8}(2)^2-ln(1)+\frac{1}{8}\\
=\frac{3}{8}+ln(2)-\frac{1}{2}+\frac{1}{8}\\
=ln(2)
\end{array}$$

$$\large\begin{array}l
30)\ L_1=-\frac{7}{2}x+5\\
L_2=-\frac{4}{5}x+5\\
L_3=x-4\\
[0,2,5]\\
\int_{0}^{2}(-\frac{4}{5}x+5+\frac{7}{2}x-5)dx+\\
\int_{2}^{5}(-\frac{4}{5}x+5+x+4)dx\\
=\left.-\frac{2}{5}x^2+\frac{7}{4}x^2\ \right|_{0}^{2}\left.-\frac{2}{5}x^2+\frac{x^2}{2}+9x\right|_{2}^{5}\\
=\frac{27}{20}-0+\frac{1}{10}(5)^2+9(5)-\frac{1}{10}(2)^2-9(2)\\
=\frac{27}{20}+\frac{5}{2}+45-\frac{2}{5}-18\\
=\frac{27}{20}+\frac{21}{10}+27\\
=\frac{69}{20}+\frac{540}{20}\\
=609/20 = 30.45



\end{array}$$

amistre64 · Answer

$$\large\begin{array}l
30)\ L_1=-\frac{7}{2}x+5\\
L_2=-\frac{4}{5}x+5\\
L_3=x-4\\
[0,2,5]\\
\int_{0}^{2}(\frac{27}{10}x)dx+\\
\int_{2}^{5}(\frac{1}{5}x+9)dx\\
=\left.\frac{27}{20}x^2\right|_{0}^{2}+\left.\frac{1}{10}x^2+9x\right|_{2}^{5}\\
=\frac{27}{20}(2)^2-0+\frac{1}{10}(5)^2+9(5)-\frac{1}{10}(2)^2-9(2)\\
=\frac{27}{5}+\frac{5}{2}+45-\frac{2}{5}-18\\
=5+\frac{5}{2}+27\\
=\frac{5}{2}+32\\

=\frac{69}{2}



\end{array}$$