hi, I am trying to solve this equation and i would like some help.
i've done some of it already and i don't know how to go on from here.
z=-\frac{(4-4i)(\sqrt{6}-i\sqrt{2})}{i}\\ =-\frac{(4-4i)(\sqrt{6}-i\sqrt{2})-i}{i(-i)}=(4i+4)(\sqrt{6}-i\sqrt{2})\\ \hspace{6}...
i have a new problem, I've tried to solve it but i think i need some advice.
\int_{0}^{\pi /2}sin7xcos6x\hspace{6} dx\\ \\ f=cos6x\hspace{6} F=\frac{sin6x}{6}\\ g'=7cos7x\hspace{6} g=sin7x\\
and this is what i have done so far
\int_{0}^{\pi /2}sin7xcos6x\hspace{6} dx\\ \\...
i think i need to see a complete solution to this problem, I've been working on this for three days now and stiill don't understand it completely.
\int_{1}^{2}(9x+6)\sqrt{7-3x} dx\
(2x+7)squareroot(7-x)dx from x=6 to x=7
u=squareroot(7-x)
u^2=(7-x)
x=7-u^2
dx=-2 du
(2x+7)squareroot(7-x)dx
(2(7-u^2)+7)u dx
(14-2u^2+7)u dx
(21-2u^2)u * -2u
(21-2u^3)*-2u
dont know how to continue from here.
(2x+7)squareroot(7-x)dx from x=6 to x=7
i've been trying to solve this integral with the integration by parts formula but it won't work.
would appreciate some help.
would like to see the whole solution and not just the answer. (the answer should bee in fractions)