Question

how do i do: S sinxcosx.dx (S is the integral sign)?

Answer 1

\[2\sin u\cos v=\sin(u+v)+\sin(u-v)\]

Answer 2

does that help?

Answer 3

I have an answer of 1/2sin^2x + C from the memo. I have no clue how it got it besides knowing i must let u=sinx

Answer 4

\[2\sin x\cos x=\sin2x\]
\[\sin x\cos x=\frac{\sin2x}{2}\]
may be nw u can integrate

Answer 5

since you wanted to use \( u\)-substitution\[\int \sin x \cos x\,\mathrm dx\\
=\frac12\int \sin 2x \,\mathrm dx\\\qquad\qquad\text{let }2x=u\\\qquad\qquad2\,\mathrm dx=\mathrm du\\\qquad\qquad\quad\mathrm dx=\frac{\mathrm du}2\\=\frac12\int \sin u\,\frac{\mathrm du}2\\=\frac14\int \sin u\,\mathrm du\]

Answer 6

oh, actually if you want to use u=sinx,

\[\int \sin x \cos x\,\mathrm dx\\
\qquad\qquad\text{let }\sin x=u\\
\qquad\qquad\,\cos x\,\mathrm dx=\mathrm du\\
=\int u\,\mathrm du\\
=\\
=
\]

Answer 7

\[\int\limits_{}^{}sinxcosx.dx=\]
let u=sinx
dx=du/cosx
therfore
\[\int\limits_{}^{}u.du= u^2/2= \frac{ 1 }{ 2 } \sin ^{2}x +C\]

Got it :)

Answer 8

great stuff!! \[\boxed{\huge\color{red}\checkmark}\]
______________


;
if you continues on from the method I was originally going to use \[=\frac12\int \sin 2x \,\mathrm dx\\=\frac14\int \sin u\,\mathrm du\\=-\frac14 \cos u+c\\=-\frac14 \cos 2x+c\].

There is some trig-identity \(\cos(2x)=1-2\sin^2x\)

\[=-\frac14 (1-2\sin^2x)+c\\=\frac{\sin^2x}2+c-\frac14\]

now \(c-\dfrac14=C\)

and we see that these two different methods yield the same result.