Question

Find the indefinite integral

Answer 1

\[\int\limits 2 \sin x + 4 \cos x dx \]

Answer 2

\[\int\limits \sin x = -\cos x \]

Answer 3

first
what function can you take the derivative of that will give you sin(x)
that (?)'=sin(x)

Answer 4

you know (cos(x))'=-sin(x)
so (-cos(x))'=?

Answer 5

ok right

Answer 6

\[\int\limits_{}^{}(2 \sin(x)+4 \cos(x)) dx \\ 2 \int\limits_{}^{}\sin(x) dx+4 \int\limits_{}^{}\cos(x) dx \\ 2(-\cos(x))+4\int\limits_{}^{}\cos(x) dx\]

Answer 7

now you just neeed to think what function can you differentiate that will give you cos(x)
that is
( ? )'=cos(x)

Answer 8

sin x ?

Answer 9

yep so the integral of cos(x) w.r.t x is sin(x)
since the derivative of sin(x) is cos(x)

Answer 10

so 2- cos x +4 sin x ?

Answer 11

\[\int\limits\limits_{}^{}(2 \sin(x)+4 \cos(x)) dx \\ 2 \int\limits\limits_{}^{}\sin(x) dx+4 \int\limits\limits_{}^{}\cos(x) dx \\ 2(-\cos(x))+4\int\limits\limits_{}^{}\cos(x) dx \\ 2(-\cos(x))+4\sin(x)+C \\ -2 \cos(x)+4\sin(x)+C\]

Answer 12

2(-cos(x))
isn't 2-cos(x)
but it is 2 times (-cos(x))

Answer 13

and since multiplication is commutative we can change the order
so instead of seeing 2(-1cos(x))
we can see (-1)(2)cos(x) or -2cos(x)

Answer 14

oh Ok!! not so bad after all

Answer 15

nope not at all

Answer 16

thank you !!

Answer 17

np :)