Question

the integration of :sin x dx÷(3sinx+sin3x)

Answer 1

Answer:

1/2√3 * tan⁻¹(tanx/√3) + c

Step-by-step explanation:

Hi,

∫ $\frac{sinx}{3sinx + sin3x}$ dx

= ∫ $\frac{sinx}{3sinx + 3sinx - 4(sinx)^{3}}$ dx

= ∫ $\frac{sinx}{6sinx - 4(sinx)^{3}}$ dx

=  ∫ $\frac{1}{6 - 4(sinx)^{2}}$ dx

Dividing Numerator and Denominator by (cosx)², we get

∫ $\frac{(secx)^{2}}{6(secx)^{2} - 4(tanx)^{2}}$ dx

Let us put t = tanx, on differentiating we get

dt = (secx)²dx    on substituting we get,

∫ $\frac{1}{6 + 6t^{2} - 4t^{2}}$ dt

= ∫ $\frac{1}{6 + 2t^{2}}$ dt

= 1/2 *  ∫ $\frac{1}{3 + t^{2}}$ dt

=1/2√3 * tan⁻¹(t/√3) + c, (c an arbitrary constant)

= 1/2√3 * tan⁻¹(tanx/√3) + c

Hope, it helped !