integrate sin x/ sin(x-a)?

Answer 1

Let I = integral of [sin(x)/sin(x-a)].dx

Substitute...
x-a = t
dx = dt

So,

I = int [sin(a+t)/sin(t)].dt

Using the formula, sin(A+B) = sinAcosB + cosAsinB
I = int { [sin(a) cos(t) + cos(a) sin(t)]/sin(t) } .dt

Splitting into 2 separate terms,
I = sin(a).int [ cot(t) ].dt + cos(a).int [ 1 ].dt

Using the formulae of integration:
int [ cot(x) ].dx = log | sin(x) | + C
int [ x^n ].dx = { x^(n+1) } / (n+1)

I = sin(a).log | sin(t) | + cos(a).t + C

Substituting the value of t back as (x-a)
I = sin(a) . log | sin(x-a) | + cos(a) . (x-a) + C

.

Answer 2

Integral (sinx / sin(x - a)] dx

Let's start off by doing substitution.

Let u = x - a.
u + a = x, therefore
du = dx

This makes our integral

Integral ( sin(u + a) / sin(u) ) du

Let's apply the sine addition trig identity on sin(u + a)

Integral ( [sin(u)cos(a) + sin(a)cos(u)] / [sin(u)] du)

Now, let's split this into two fractions and two integrals.

Integral ( [sin(u)cos(a)] / [sin(u)] ) du + Integral
( [sin(a)cos(u)] / sin(u) ) du

Notice that some stuff cancels, leaving us with

Integral (cos(a)) du + Integral (sin(a) [cos(u)/sin(u)]) du

Note that anything with an "a" in it is a constant and can be factored out of the integral.

cos(a) * Integral (1) du + sin(a) * Integral ([cos(u)/sin(u)]) du

Now, we can solve these integrals easily. Let's solve the second integral aside by substitution.

Integral (cos(u)/sin(u)) du

Let v = sin(u). dv = cos(u) du, so we have

cos(a) * Integral (1) du + sin(a) * Integral ([1/v]) dv

cos(a) [u] + sin(a) ln|v| + C

Now, we substitute v = sin(u).

cos(a) [u] + sin(a) ln|sin(u)| + C

Now, we resubstitute u = x - a

cos(a) [x - a] + sin(a) ln |sin (x - a)| + C

Answer 3

let x - a = t
dt = dx

sin x/ sin(x-a) = sin (t + a)/ sin t = [sin t cos a + cos t sin a ]/sint

=cos(a) +sin(a) cot(t)

integrating we get

therefore integral of sin x/ sin(x-a) = t cos(a) + sin(a) ln{sin(t)} + K
where K is constant of integration

or (x - a) cos (a) + sin(a) ln{sni(x - a)} + K

i hope it will help

Answer 4

put x-a=z
x=a+z
int[sin x/sin x-a]
=int[sin(a+z)/sin z
=int[sina cos z+cos a sinz]/sin z
=int[sin a cotz+cos a]
=sin a ln sin z+z cos a
=sin a ln sin[x-a]+cos a [x-a]

Answer 5

sinx / sin(x-a)
= sin(x-a+a) / sin(x-a)
= [sin(x-a)cosa + cos(x-a)sina] / sin(x-a)
= cosa + cotg(x-a)sina

The integral is
x*cosa + sina*ln|sin(x-a)| + C.

Openpsychy and Laeeq (below) also have it, we only differ by
-a*cosa which can be absorbed into the constant term.

Answer 6

use the quotient rule: f(x)=u/v f'(x)=(u'v-v'u)/(2v)
u=sin(x), u'=cos(x), v=sin(x-a), v'=cos(x-a)
f'(x)=(cos(x)sin(x-a)-sin(x)cos(x-a))/(2sin(x-a))

Answer 7

answer:
sin(a)*
ln(tan(x)*cos(a)-sin(a))-
1/2*sin(a)*ln(tan(x)^2+1)+
cos(a)*arctan(tan(x))