Question

Alright guys Trivial question!

Find the Definite Integral of Sin^7x dx

Answer 1

Indefinite*

Answer 2

doesn't look trivial to me...

Answer 3

it's just a lot of trig transformations =_=

Answer 4

Hey, it's fun:)

Answer 5

be my guest... without using Mr. Wolfram...

Answer 6

Integration by parts.

Answer 7

Here, take u = sin^6 x and dv= sinxdx

Answer 8

\[du= 6\sin^5xcos x dx\]

Answer 9

v= -cosx

Answer 10

so you're integral =
\[-cosxsin^6 x - \int\limits_{}^{} -6cosxsin^5xcosxdx\]

Answer 11

\[= -\sin^6xcosx + 6\int\limits_{}^{} \sin^5x(\cos^2x)dx\]

Answer 12

\[= -\sin^6xcosx +6\int\limits_{}^{} \sin^5x (1-\sin^2x) dx\]

Answer 13

\[\int\limits_{}^{}\sin^7dx= -\sin^6xcosx + 6\int\limits_{}^{} \sin^5x - \sin^7x dx\]

Answer 14

Break up the two integrals on the right hand side of the equal sign

Answer 15

\[ \int\limits_{}^{} \sin^7xdx = -\sin^6xcosx + 6\int\limits_{}^{} \sin^5xdx - 6\int\limits_{}^{} \sin^7x dx\]

Answer 16

Notice we have two like terms? Put them together to get

Answer 17

\[7\int\limits_{}^{}\sin^7dx = -\sin^6xcosx+ 6\int\limits_{}^{}\sin^5x\]

Answer 18

Divide everything by seven to get

Answer 19

\[\int\limits_{]}^{}\sin^7x dx= -1/7\sin^6xcosx+6/7\int\limits_{}^{}\sin^5x\]

Answer 20

And that's your answer. Trivia!!

Answer 21

I don't even know who Mr. Wolfram is.

Answer 22

i wonder if u can do this with the tabular method to continue on to do that other integral?

Answer 23

I know that form is accepted, and you don't have to go on finding the integral. But in som it may be easy so one can do it.

Answer 24

good work man...:)

Answer 25

Haha thanks, it's quite long I must say.

Answer 26

YA and i guess it's even longer when you use substitution method

Answer 27

Definitely wouldn't want to try that, but there's guidelines for when to use trig substitutions

Answer 28

can u tell me what is that?

Answer 29

Sure

Answer 30

When dealing with trig integrals, If:
The power of cosine is odd: Save a cosine and change everything else to sine and then take u=sin x
The power of sine is odd: Save a sine and change everything else to cosine then take u=cosx
The power of cosx and sinx are both even: Use half-angle formulas.
The power of taangent is odd: In most cases, useful substitution is u= secx
The power of secant is even: Save a sec^2 x and change everything else to tangent. Take u= tan x. Those are the guidelines

Answer 31

sin^7x dx = sin^6(x) sin(x)dx

= (sin^2x)^3 sinx dx = (1-cos^2x) ^3 sinx dx

let u = cosx

du = -sinx dx

therefore, we get:

(1 - u^2) ^3 du

= 1 - 3u^2 + 3u^4 - u^5 du

integrate from there.

Answer 32

thanks
beeqay

Answer 33

actually i usually get confused in such type of stuffs

Answer 34

no problem
and @Hermeezey you forgot that it would be -du

Answer 35

oh yea sorry, I'm easily confused when I'm typing numbers on the computer

Answer 36

Meant to ask for the solution using integration by parts, because it's more challenging.