what is the integral of sinx/sin4x - Maths - Integrals

Question

what is the integral of sinx/sin4x

Lalit Mehra · Accepted Answer

$I = \int \frac{\sin x}{\sin 4 x} d x = \int \frac{\sin x}{2 \sin 2 x \cos 2 x} d x = \int \frac{\sin x}{2 (2 \sin x \cos x) \cos x} d x = \frac{1}{4} \int \frac{1}{\cos x \cos 2 x} d x = \frac{1}{4} \int \frac{\cos x}{\cos^{2} x \cos 2 x} d x = \frac{1}{4} \int \frac{\cos x}{(1 - \sin^{2} x) (1 - 2 \sin^{2} x)} d x$

Putting sin x = z and cos x dx = dz, we get

$I = \frac{1}{4} \int \frac{d z}{(1 - z^{2}) (1 - 2 z^{2})}$

Putting z ²= t, we get

$\frac{1}{(1 - z^{2}) (1 - 2 z^{2})} = \frac{1}{(1 - t) (1 - 2 t)} ∴ 1 = A (1 - 2 t) + B (1 - t)$

When t = 1, we get

– A = 1

∴ A = – 1

When $t = \frac{1}{2}$ , we get

$\frac{1}{2} B = 1 ∴ B = 2 \frac{1}{(1 - t) (1 - 2 t)} = \frac{- 1}{1 - t} + \frac{2}{1 - 2 t} ∴ \frac{1}{(1 - z^{2}) (1 - 2 z^{2})} = \frac{1}{1 - z^{2}} + \frac{2}{1 - 2 z^{2}} \Rightarrow I = \frac{1}{4} \int \frac{d z}{(1 - z^{2}) (1 - 2 z^{2})} = - \frac{1}{4} \int \frac{d z}{1 - z^{2}} + \frac{1}{2} \int \frac{d z}{1 - 2 z^{2}} \Rightarrow I = \frac{1}{4} \int \frac{d z}{(1)^{2} - z^{2}} + \frac{1}{2} \int \frac{d z}{(1)^{2} - (\sqrt{2} z)^{2}} \Rightarrow I = - \frac{1}{4} \times \frac{1}{2} \log | \frac{1 + z}{1 - z} | + \frac{1}{2} \times \frac{1}{2 \sqrt{2}} \log | \frac{1 + \sqrt{2} z}{1 - \sqrt{2} z} | + C \Rightarrow I = - \frac{1}{8} \log | \frac{1 + \sin x}{1 - \sin x} | + \frac{1}{4 \sqrt{2}} \log | \frac{1 + \sqrt{2} \sin x}{1 - \sqrt{2} \sin x} | + C$

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