Question

Another u-substitution integration.. : sin^6(5x)cos(5x)dx

Answer 1

u = 5x, du = 5dx

Answer 2

\[\int\limits_{}^{}\sin(u)^6 \cos(u) \frac{ du }{ 5 }\]

Answer 3

the answer is \[\frac{ 1 }{ 35 }\sin^7 (5x) + C\]

Answer 4

but im not sure how to eliminate the \[\cos(u)\] in the integral

Answer 5

try : \(u = \sin(5x)\)

Answer 6

alright, let's see

Answer 7

would the derivative be -cos5x?

Answer 8

\(\frac{d}{dx}\sin x = \cos x\)

Answer 9

\[\int\limits_{}^{}(u)^6 (du)\]

Answer 10

dont forget the chain rule

Answer 11

u dont have just sinx,
u are substituting sin(5x)

Answer 12

5cos(5x)

Answer 13

\(u = \sin 5x\)

\( du = 5\cos (5x) dx \)

Answer 14

\[\int\limits_{}^{}(u)^6 5du \]

Answer 15

not quite, lets do it step by step

Answer 16

alright, sorry and thanks!!

Answer 17

\(u = \sin 5x\)

\( du = 5\cos (5x) dx \)

\(\implies \)


\( \dfrac{du}5{} = \cos (5x) dx \)

Answer 18

\[\int \sin^6(5x) \cos(5x)dx \]

becomes :

\[\int u^6 \frac{du}{5} \]

Answer 19

oh, okay I understand

Answer 20

is it always safe to assume that: du = 1?

Answer 21

i got the answer btw, just wanted to know ^

Answer 22

hmm what do u mean assume `du = 1` ?

Answer 23

hm, \[\int\limits_{}^{}(u)^6 (\frac{ du }{ 5 })\] becomes

Answer 24

\[(\frac{ 1 }{ 7 }u^7) (\frac{ 1 }{ 5 }) + C \]

Answer 25

^thats right,
i think i got what you're asking

Answer 26

\(\color{red}{\int} \) is meaningless without the differential \(\color{red}{dx}\) or \(\color{red}{du}\)

Answer 27

below "entire thing" is called the integral operator :

\(\color{red}{\int~dx} \)

Answer 28

you can think of \(\color{red}{\int}\) as the "starting parenthesis", and the \(\color{red}{dx}\) or \(\color{red}{du}\) as the "ending parenthesis"

Answer 29

ah, that makes a lot of sense. Thanks so much! i would give double medal if I could!

Answer 30

And, the stuff that lies in between is called the `integrand`, which you will be integrating :

\(\color{red}{\int}\) `integrand`\(\color{red}{dx}\)

Answer 31

np... u wlc :)