Verified answer

This requires a special substitution. Let t = tan(x/2).

Then, sin x = 2t/(1+t^2), cos x = (1-t^2)/(1+t^2), and dx = 2dt/(1+t^2).

Our integral becomes

∫dx/(3-5sinx) = ∫ [2dt/(1+t^2)]/[3-5(2t/(1+t^2))]

= ∫ 2 dt/ [3(1+t^2) - 10t)]

= ∫ 2 dt/ (3t^2 - 10t + 3)

= ∫ 2 dt/ [(3t - 1)(t - 3)].

By partial fractions, we have

2 / [(3t - 1)(t - 3)] = A/(3t - 1) + B/(t - 3)

==> 2 = A(t - 3) + B(3t - 1)

Letting t = 3 gives B = 1/4.

Letting t = 1/3 gives A = -3/4

Thus, our integral becomes

(1/4) * ∫ [-3/(3t - 1) + 1/(t - 3)] dt

= (1/4) * [-ln |3t - 1| + ln |t - 3|] + C

= (1/4) * ln|(t - 3)/ (3t - 1)| + C.

With t = tan(x/2), our final answer is

(1/4) * ln|(tan(x/2) - 3)/ (3tan(x/2) - 1)| + C.

I hope that helps!

If f(x) is a rational expression in sin(x) or cos(x), the substitution 2 atan(u) = x transforms ∫ f(x) dx into the integral of a rational function of u.

∫ dx / (3 - 5sin(x))

let: 2 atan(u) = x ; (2 du) / (1 + u²) = dx

note: cos(x) = (1 - u²)/(1 + u²) ; sin(x) = 2u/(1+u²)

= ∫ 2 / (3u² - 10u + 3) du

= ∫ 2 / ((u -3)(3u - 1)) du

... 2 / ((u -3)(3u - 1)) = A / (u - 3) + B / (3u - 1)

... 2 = A (3u - 1) + B (u - 3)

... A = 1/4 ..... let u = 3

... B = -3/4 .... let u = 1/3

= ∫ du / (4(u - 3)) - 3 ∫ du / (4(3u - 1))

= 1/4 ln(u - 3) - 1/4 ln(3u - 1) + c

= 1/4 (ln(tan(x/2) - 3) - ln(3 tan(x/2) - 1)) + c

Answer: 1/4 (ln(tan(x/2) - 3) - ln(3 tan(x/2) - 1)) + c

integral 1/(3-5 sin(x)) dx = 1/4 log(3 cos(x/2)-sin(x/2))-1/4 log(cos(x/2)-3 sin(x/2))+constant