Question

How to SOLVE: medals can be given if it satisfies u
Integrate the following indefinite integral:

(sin7x)^12 * (cos7x)^3

Hint: sin^2 + cos^2 = 1

Answer 1

OK, you're NOT gonna want to do anything with it but:

{ 1/22364160 } * { 13*\sin(105x) - 135*\sin(91x) + 585*\sin(77x) - 1235*\sin(63x) + 585*\sin(49x) + 3861*\sin(35x) - 12155*\sin(21x) + 19305*\sin(7x) } + C

Answer 2

http://wims.unice.fr/wims/wims.cgi?session=EZ2B979C52.1&+lang=en&+module=tool%2Fanalysis%2Ffunction.en

Answer 3

check it yourself

Answer 4

any objections

Answer 5

This integral is why you memorize the reduction formulas. http://en.wikipedia.org/wiki/Integration_by_reduction_formulae

Answer 6

it was just how to solve

Answer 7

initially hows that

Answer 8

You could also do it by painfully redefining each\[\sin^2x=1-\cos^2x\]and its reverse alternative. If you run into a problem, I think dividing all sides by \(\cos^2x\) for an easy \(\sec^2x\to\tan x\) relation would be the best bet.

Answer 9

how about turning \[(\cos 7x)^3 \implies (1 - \sin^2 7x)(\cos 7x)\]

Answer 10

\[\int (\sin^{12} 7x - \sin^{14} 7x)(\cos x)dx\]

Answer 11

now it's u-sub

Answer 12

okay

Answer 13

then

Answer 14

lol should i have to complete it =))

Answer 15

no medal me if my answer satisfies u

Answer 16

\[
(sin7x)^{12} * (cos7x)^3=(sin7x)^{12} * (cos7x)^2 (cos7x) =
(sin7x)^{12} (1- (sin7x))^2 (cos7x)
\]

Answer 17

u= sin( 7x) and the rest is easy.

Answer 18

\[\frac 17\int \sin^{12} u)(\cos u) du - \frac 17 \int (\sin^{14} u )(\cos u du\]

Answer 19

You probably know this, but use \(\text{\\}\) in \(\TeX\) if you want a line break.

Answer 20

as u r an retired professer

Answer 21

\[\frac{\sin^{13}u}{91} - \frac{\sin^{15} u}{105}\]

Answer 22

+C

Answer 23

okay

Answer 24

hmmmmmmmmmm

Answer 25

happy now lol

Answer 26

@eliassaab
I saw u r website it helped me a lot

Answer 27

Thanks @Rohangrr

Answer 28

@eliassaab you wrote cos7x=1-sin7x...is it true?

Answer 29

yes its

Answer 30

so roh...

Answer 31

Sorry
\[ 1 - \sin^2(7x) \]

Answer 32

okay

Answer 33

lgb

Answer 34

lol still not happy?

Answer 35

@Rohangrr give me a medal man!!

Answer 36

why soooooo

Answer 37

it's obviously right

Answer 38

elisaab deserved the medal from my side

Answer 39

i found a big mistake if not ur ans will be wrong

Answer 40

okay

Answer 41

lol whatever...what i want is you to admit i was right

Answer 42

wait got it

Answer 43

\[

(sin7x)^{12} * (cos7x)^3=(sin7x)^{12} * (cos7x)^2 (cos7x) =\\
(sin7x)^{12} (1- (sin7x)^2) (cos7x)
\]

Answer 44

well I am running behind the answer

Answer 45

or running behind the answer

Answer 46

@lgbasallote
i think u made a mistake
\[(\sin 7x)^{12} \]
not
\[\sin^{12} (7x)\]

Answer 47

\[(\sin 7x)^12 \implies \sin^{12} 7x \cdots\]

Answer 48

^you now what i mean lol

Answer 49

I dislike the notation as well, since it isn't internally consistent with\[\arcsin x=\sin^{-1}x\neq\frac1{\sin x}\]

Answer 50

@lgbasallote
sorry
Oh, what did I say
my mistake
;)

Answer 51

i made no mistake :) @Rohangrr just refuses to admit im right haha