Question

Integrate :

Answer 1

\[\Huge \int\limits \frac{\sin x}{\sin 4x}\]

Answer 2

by parts i assume

Answer 3

didn't work

Answer 4

sin4x=2sin2xcos2x=4sinxcosxcos2x
\[\Huge \int\limits \frac{\cancel{\sin x}}{4 \cancel{\sin x} \cos x \cos 2x}= \int\limits \frac{dx}{\cos x \cos 2x}\]

Answer 5

also 1/4

Answer 6

http://www.wolframalpha.com/input/?i=integral+of+sin%28x%29+%2F+sin%284x%29

Answer 7

\[\Huge \frac{ 1}{4} \int\limits \frac{dx}{ \cos x \cos 2x}\]

Answer 8

this seems very hard by the answer they gave...

Answer 9

still want it?

Answer 10

if there's a simpler way

Answer 11

subsitution maybe

Answer 12

\[Cos(2x) = \cos^2(x)+\sin^2(x) = 1-2\sin^2(x)=2\cos^2(x)-1\]

Answer 13

another way to do it would be to use a trig substitution
since sin (4x) = 2 sin (2x) cos (2x)

Answer 14

This one is gross

Answer 15

\[\frac{ 1 }{ \sin4x } = \csc4x\] try

Answer 16

sinxcsc4x = ?

Answer 17

http://www.wolframalpha.com/input/?i=sinxcsc4x

\[\int\limits \frac{ 1 }{ 2\cos^3x+2cosx-6\sin^2xcosx } dx\]

Answer 18

This will get you to the answer, but will take you a while...

Answer 19

@Kainui need ure halp plz

Answer 20

Sorry I was afking

Answer 21

My first thought is to make the substitution 4x=u so that you're messing with the numerator instead of the denominator.

Answer 22

but idk lol

Answer 23

Don't think that works idk

Answer 24

Are we allowed to use complex numbers?

Answer 25

I guess so

Answer 26

\[\int\limits\limits \frac{ 1 }{ 2\cos^3x+2cosx-6\sin^2xcosx } dx \times \frac{ cosx }{ cosx }\]

Answer 27

U sub then partial fractions...omg im out

Answer 28

@DLS are there bounds on this integral? It might be easy to solve by making symmetry arguments.

Answer 29

nope no bounds , the next question however has bounds from 0 to 1

Answer 30

Let \[I = \int \frac{sin _\ x}{sin _\ 4x} dx \]
So \[ I = \int \frac{cosx _\ dx}{4cos^2 x _\ cos_\ 2x}\]
\[I = \int \frac{cos_\ x _\ dx}{4(1-sin^2 x)(1-2sin^2 x)}\]
Now substitute sin x = u so that
\[I = \int \frac{du}{4(1-u^2 )(1-2u^2)}\]
Now use partial fractions to integrate with respect to u and then inverse substitution, u = sin x.
\[I = \frac{1}{4}\int \left( \frac{2}{(1-2u^2)} - \frac{1}{(1-u^2)} \right) du \]
\[I = \frac{1}{8}\left( 2\ln \frac{\sqrt {2} u + 1}{(\sqrt{2}u - 1 )} -\ln \frac{1+u}{(1-u)} \right) + C\]
\[I = \frac{1}{8}\left( 2\ln \frac{\sqrt {2} sin_\ x+ 1}{\sqrt{2}sin_\ x- 1 } -\ln \frac{1+sin_\ x}{1-sin_\ x} \right) + C\]

Hope this helps.

Alternatively, You can use complex numbers and De Moivre's theorem.
You can use \[ sin _\ x = \frac{\exp (i x) - \exp(-i x)}{2i} \]
and
\[ sin _\ 4x = \frac{\exp (4i x) - \exp(-4i x)}{2i} \]