Question

integrate sin(x)/sin(4x)

Answer 1

how about opening every term then solve it you must have done that

Answer 2

Now when you open the denominator

\[\frac{1}{4}\int\limits \frac{1}{cosx(2\cos^2x -1)}dx\]

Answer 3

i tried that i could solve upto the first four expressions in this no idea for how to get the expression with (i) http://integrals.wolfram.com/index.jsp?expr=sin%28x%29%2Fsin%284x%29&random=false

Answer 4

thats too long

Answer 5

why do you want to solve it ...i hate lengthy problems in calculus

Answer 6

sin(4x) = 2sin(2x)cos(2x) = 4sin(x)cos(x)cos(2x)

also,
cos(2x) = 1-2sin^2(x)
so

sin(4x)/sin(x) = 4cos(x)cos(2x)
= 4cos(x) - 8cos(x)sin^2(x)

Answer 7

So first, remember the identity that 1/sin(4x) = csc(4x)
Now use the product rule.
csc(4x)*cos(x) + 4(-csc(x) cot(x)) * sin(x)
So with some manipulation, your final answer is:
cos(x)csc(4x)-4sin(x)cot(4x)csc(4x)

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