Question

evaluate eschmaluate ....

Answer 1

homeworks :)

Answer 2

\[\Large
\begin{array}c
4)\ \int_{0}^{2\sqrt{3}}\frac{x^3}{\sqrt{16-x^2}}dx
\end{array}
\]

Answer 3

\[\Large
\begin{array}c
6)\ \int_{1}^{2}\frac{\sqrt{x^2-1}}{x}dx
\end{array}\]

Answer 4

\[\Large
\begin{array}c
10)\ \int\frac{t^5}{\sqrt{t^2+2}}dt
\end{array}\]

Answer 5

\[\Large
\begin{array}c
12)\ \int_{0}^{1}{x}{\sqrt{x^2+4}}\ dx
\end{array}\]

Answer 6

\[\Large
\begin{array}c
13)\ \int\frac{\sqrt{x^2-9}}{x^3}\ dx
\end{array}\]

Answer 7

\[\Large
\begin{array}c
14)\ \int\frac{1}{u\sqrt{5-u^2}}\ du
\end{array}\]

Answer 8

what fun...

Answer 9

\[\Large
\begin{array}l
4)\ \int_{0}^{2\sqrt{3}}\frac{x^3}{\sqrt{16-x^2}}dx\\\\
= \int_{0}^{2\sqrt{3}}\frac{4^3sin^3(t)}{4cos(t)}4cos(t)dt\\\\
= 4^3\int_{0}^{2\sqrt{3}}sin^3(t)dt\\\\
= 4^3\int_{0}^{2\sqrt{3}}(1-cos^2(t))sin(t)dt\\\\
= 4^3\int_{0}^{2\sqrt{3}}sin(t)-sin(t)cos^2(t)dt\\\\
= 4^3(-cos(t)+\frac{cos^3(t)}{3});[0,2\sqrt{3}]\\\\
= -4^2\sqrt{16-x^2}+\frac{\sqrt{(16-x^2)^3}}{3}\\\\
= -4^2\sqrt{16-(2\sqrt{3})^2}+\frac{\sqrt{(16-(2\sqrt{3})^2)^3}}{3}\\\\
+4^2\sqrt{16-0^2}-\frac{\sqrt{(16-0^2)^3}}{3}\\\\
= -32+\frac{{8}}{3} +64-\frac{64}{3}\\\\
= 32-\frac{56}{3}=\frac{40}{3}\\\\

\end{array}\]

Answer 10

\[\Large

\begin{array}l
6)\ \int_{1}^{2}\frac{\sqrt{x^2-1}}{x}dx\\\\
=\int_{1}^{2}\frac{tan(t)}{sec(t)}sec(t)tan(t)dt\\\\
=\int_{1}^{2}tan^2(t)dt\\\\
=tan(t)-t;[1,2]\\\\
=\sqrt{x^2-1}-sec^{-1}(x)\\\\
=\sqrt{2^2-1}-sec^{-1}(2)\\\\
-\sqrt{1^2-1}+sec^{-1}(1)\\\\
=\sqrt{3}-\frac{\pi}{3}\\\\

\end{array}\]

Answer 11

\[\Large

\begin{array}l
10)\ \int\frac{x^5}{\sqrt{x^2+2}}dx\\\\
=\int\frac{\sqrt{2}^5tan^5(t)}{\sqrt{2}sec(t)}\sqrt{2}sec^2(t)dt\\\\
=\sqrt{2}^5\int{tan^5(t)}sec(t)dt\\\\
=\sqrt{2}^5\int{tan^4(t)}(tan(t)sec(t))dt\\\\
=\sqrt{2}^5\int{(sec^4(t)-2sec^2(t)+1))}\\\\(tan(t)sec(t))dt\\\\
=\sqrt{2}^5( \frac{sec^5(t)}{5}-\frac{2sec^3(t)}{3}+sec(t)\\\\
=\sqrt{2}^5( \frac{\sqrt{2+x^2}^5}{5.\sqrt{2}^5}+\frac{2\sqrt{2+x^2}^3}{3.\sqrt{2}^3}+\frac{\sqrt{2+x^2}}{\sqrt{2}})\\\\
=\frac{\sqrt{2+x^2}^5}{5}-\frac{4\sqrt{2+x^2}^3}{3}+4\sqrt{2+x^2}\\\\
=\sqrt{2+x^2}(\frac{\sqrt{2+x^2}^4}{5}-\frac{4\sqrt{2+x^2}^2}{3}+4)\\\\
=\sqrt{2+x^2}(\frac{4+4x^2+x^4}{5}-\frac{8+4x^2}{3}+4)\\\\
=\sqrt{2+x^2}\\\\
(\frac{12+12x^2+3x^4-40-20x^2+60}{15})\\\\
=\sqrt{2+x^2}(\frac{32-8x^2+3x^4}{15})\\\\
=\frac{\sqrt{2+x^2}}{15}(3x^4-8x^2+32)\\\\
\end{array}\]

Answer 12

\[\Large
\begin{array}l
12)\ \int_{0}^{1}{x}{\sqrt{x^2+4}}\ dx\\\\
=6\int_{0}^{1} sec^2(t)(tan(t)sec(t))dt\\\\
=6\frac{sec^3(t)}{3};[0,1]\\\\
=\frac{\sqrt{x^2+4}^3}{4}-\frac{\sqrt{x^2+4}^3}{4}\\\\
=\frac{\sqrt{1^2+4}^3}{4}-\frac{\sqrt{0^2+4}^3}{4}\\\\
=\frac{5\sqrt{5}-8}{4}
\end{array}\]

Answer 13

\[\Large
\begin{array}l
13)\ \int\frac{\sqrt{x^2-9}}{x^3}\ dx\\\\
=\int\frac{3tan(t)}{3^3sec^3(t)}\ 3sec(t)tan(t)dt\\\\
=\int\frac{tan^2(t)}{3sec^2(t)}\ dt\\\\
=\frac{1}{3}\int sin^2(t)\ dt\\\\
=\frac{1}{3}(-\frac{sin(t)cos(t)}{2}-\frac{t}{2})\\\\
=\frac{1}{3}(-\frac{3\sqrt{x^2-9}}{2x^2}-\frac{sec^{-1}(\frac{x}{3})}{2})\\\\
=-\frac{\sqrt{x^2-9}}{2x^2}-\frac{1}{6}tan^{-1}(\frac{x}{3})+C\\\\



\end{array}\]

Answer 14

\[\Large
\begin{array}l
7)\ \int\frac{x}{x-6}\ dx=\int\frac{x+(6-6)}{x-6}\ dx\\\\
=\int\frac{x-6}{x-6}+\frac{6}{x-6}\ dx\\\\
\\\\
=x+6\ ln|x-6|+C\\\\

9)\ \int\frac{x-9}{(x+5)(x-2)}\ dx\\\\
=\int\frac{A}{x+5}dx+\int\frac{B}{x-2}\ dx\\\\
=A\ ln|x+5|+B\ ln|x-2|+C\\\\
x-9=A(x-2)+B(x+5)\\\\
x=-5;\ A=2:x=2;\ B=-1\\\\
=2\ ln|x+5|-ln|x-2|+C\\\\

10)\ \int\frac{1}{(x+4)(x-1)}\ dx\\\\
=\int\frac{A}{x+4}dx+\int\frac{B}{x-1}\ dx\\\\
=A\ ln|x+4|+B\ ln|x-1|+C\\\\
1=A(x-1)+B(x+4)\\\\
x=-4;\ A=-\frac{1}{5}:x=1;\ B=\frac{1}{5}\\\\
=-\frac{1}{5}(ln|x+4|-ln|x-1|)+C\\\\

11)\ \int\frac{1}{x^2-1}\ dx\\\\
=\int\frac{A}{x-1}dx+\int\frac{B}{x+1}\ dx\\\\
=A\ ln|x-1|+B\ ln|x+1|+C\\\\
1=A(x+1)+B(x-1)\\\\
x=1;\ A=\frac{1}{2}:x=-1;\ B=-\frac{1}{2}\\\\
=\frac{1}{2}(ln|x-1|-ln|x+1|)+C\\\\

12)\ \int\frac{x-1}{x^2+3x+2}\ dx\\\\
=\int\frac{A}{x+2}dx+\int\frac{B}{x+1}\ dx\\\\
=A\ ln|x+2|+B\ ln|x+1|+C\\\\
x-1=A(x+1)+B(x+2)\\\\
x=-2;\ A=3:x=-1;\ B=-2\\\\
=3ln|x+2|-2ln|x+1|+C\\\\


\end{array}\]

Answer 15

\[\Large
\begin{array}l
7)\ \int\frac{x}{x-6}\ dx=\int\frac{x+(6-6)}{x-6}\ dx\\\\
=\int\frac{x-6}{x-6}+\frac{6}{x-6}\ dx\\\\
\\\\
=x+6\ ln|x-6|+C\\\\

9)\ \int\frac{x-9}{(x+5)(x-2)}\ dx\\\\
=\int\frac{A}{x+5}dx+\int\frac{B}{x-2}\ dx\\\\
=A\ ln|x+5|+B\ ln|x-2|+C\\\\
x-9=A(x-2)+B(x+5)\\\\
x=-5;\ A=2:x=2;\ B=-1\\\\
=2\ ln|x+5|-ln|x-2|+C\\\\

10)\ \int\frac{1}{(x+4)(x-1)}\ dx\\\\
=\int\frac{A}{x+4}dx+\int\frac{B}{x-1}\ dx\\\\
=A\ ln|x+4|+B\ ln|x-1|+C\\\\
1=A(x-1)+B(x+4)\\\\
x=-4;\ A=-\frac{1}{5}:x=1;\ B=\frac{1}{5}\\\\
=-\frac{1}{5}(ln|x+4|-ln|x-1|)+C\\\\

11)\ \int_{2}^{3}\frac{1}{x^2-1}\ dx\\\\
=\int\frac{A}{x-1}dx+\int\frac{B}{x+1}\ dx\\\\
=A\ ln|x-1|+B\ ln|x+1|+C\\\\
1=A(x+1)+B(x-1)\\\\
x=1;\ A=\frac{1}{2}:x=-1;\ B=-\frac{1}{2}\\\\
=\frac{1}{2}(ln|x-1|-ln|x+1|);\ [2,3]\\\\
=\frac{1}{2}(ln|3-1|-ln|3+1|)\\\\
-\frac{1}{2}(ln|2-1|-ln|2+1|)\\\\
=\frac{1}{2}(ln(2)-ln(4))-\frac{1}{2}(ln(1)-ln(3))\\\\
=\frac{1}{2}ln(\frac{1}{2})-\frac{1}{2}ln(\frac{1}{3})\\\\
=\frac{1}{2}ln(\frac{3}{2})\\\\

12)\ \int_{0}^{1}\frac{x-1}{x^2+3x+2}\ dx\\\\
=\int\frac{A}{x+2}dx+\int\frac{B}{x+1}\ dx\\\\
=A\ ln|x+2|+B\ ln|x+1|+C\\\\
x-1=A(x+1)+B(x+2)\\\\
x=-2;\ A=3:x=-1;\ B=-2\\\\
=3ln|x+2|-2ln|x+1|;\ [0,1]\\\\
=3ln|1+2|-2ln|1+1|\\\\
-3ln|0+2|+2ln|0+1|\\\\
=3ln|3|-2ln|2|-3ln|2|+2ln|1|\\\\
=ln|3^3|-ln|2^2|-ln|2^3|\\\\
=ln(\frac{27/4}{8})
=ln(\frac{27}{32})



\end{array}\]