Hi there,
Urgently need help with the following questions!
1.Integral of cos^8xsin^3xdx
2.Integral of (x^2 +6x)dy=y^2dx-12dy
3.Determinant of:
3-λ -1 2
5 -3-λ 5
1 -1 2-λ
Thanks so much if you can help! xxx
Hi there,
Urgently need help with the following questions!
1.Integral of cos^8xsin^3xdx
2.Integral of (x^2 +6x)dy=y^2dx-12dy
3.Determinant of:
3-λ -1 2
5 -3-λ 5
1 -1 2-λ
Thanks so much if you can help! xxx
$\displaystyle \begin{gathered}
= \int {} cos^8 (x)sin^2 (x)sin(x)dx \hfill \\
= \int {} cos^8 (x)(1 - \cos ^2 (x))sin(x)dx \hfill \\
\end{gathered}
$
$\displaystyle \begin{array}{|c|}
\hline SUBSTITUTION\\
\hline a = \cos (x) \\
a~da = - \sin (x)dx \\
- a~da = \sin (x)dx \\\hline
\end{array}$
$\displaystyle = \int a^8(1 - a^2)da$
$\displaystyle = \int (a^8 - a^{10})da$
$\displaystyle = \frac {a^9}{9} - \frac{a^{11}}{11}+C$
Unsubstitute
$\displaystyle = \frac {cos^9(x)}{9} - \frac{cos^{11}(x)}{11}+C$
Hello, Confuzzled!
2. Integral of $\displaystyle (x^2 +6x)dy \:=\:y^2dx-12dy$
We have: .$\displaystyle (x^2 + 6x)dy + 12dy \:=\:y^2dx$
. . . . . . . . . $\displaystyle (x^2 + 6x + 12)dy \;=\;y^2dx$
Separate variables: .$\displaystyle \frac{dy}{y^2} \;=\;\frac{dx}{x^2+6x+12} $
Integrate: .$\displaystyle \int y^{-2}dy\;=\;\int\frac{dx}{(x+3)^2 + (\sqrt{3})^2} $
. . and we have: .$\displaystyle -y^{-1} \;=\;\frac{1}{\sqrt{3}}\arctan\left(\frac{x+3}{\sq rt{3}}\right) + C$
I'll let you simplify it . . .