Question

hlo.. do it step by step..​

Answer 1

$I = \int \: xsinxsin2xsin3x \\ or \\I =\int \: xsin2xsinxsin3x \\sin \alpha sin \beta = \frac{1}{2} |cos( \alpha - \beta ) - cos ( \alpha + \beta )| \\ using \: this \: formula \: we \: can \: write \\ I=\int \:xsin2x \times \frac{1}{2} |cos(x - 3x) - cos(x + 3x)| \\ or \\ = \frac{1}{2} \int \: xsin2xcos2x - xsin2xcos4x \\ using \: integration \: by \: part \: we \: can \: write \\ I = \frac{1}{2} \int \: xsin2xcos2x - \frac{1}{2} \int \: xsin2xcos4x \\ take \: I1 = \frac{1}{2}\int \: xsin2xcos2x \\ using \: formula \: \sin\alpha \cos \alpha = \frac{1}{2} \sin2 \alpha \\ I1 = \frac{1}{4}\int \: xsin4 \alpha \\ now \: using \: formula \: of \: integration \: by \: part \\ \int\: udv = uv - \int \: udv \\ I1 = \frac{1}{4} | \frac{ - xcos4x}{4} - \int \frac{ - cos4x}{4}dx| \\ = \frac{1}{4} | \frac{ - xcos4x}{4} + \frac{1}{4} \int \: cos4x| \\ I1 = \frac{sin4x}{64} - \frac{xcos4x}{16} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1) \\ let \: I2 = \frac{1}{2} \int \: xsin2xcos4x \\ sin \alpha cos \beta = \frac{1}{2} |sin( \alpha + \beta ) + sin( \alpha - \beta ) | \\ I2 = \frac{1}{2} \int \: x \times \frac{1}{2} |sin(4x + 2x) - sin(4x - 2x)| \\ = \frac{1}{4} \int \:xsin6x - \frac{1}{4} \int \: xsin2x \\ I21 = \frac{1}{2}\int \: xsin6x \\ I21 = \frac{1}{4}| \frac{ - xcos6x}{6} - \int \frac{ - cos6x}{6}dx | \\ I21 = \frac{1}{4} | \frac{ - xcos6x}{6} + \frac{sin6x}{36} | \\ I21 = \frac{sin6x}{144} - \frac{xcos6x}{24} \\ I22 = \frac{1}{4} \int \: xsin2x \\ again \: using \: integration \: by \: part \: rule \\ we \: can \: write \\ I22 = \frac{1}{4} | \frac{ - xcos2x}{2} - \int \frac{ - cos2x}{2}dx| \\ = \frac{sin2x}{16} - \frac{xcos2x}{8} \\ I2=I21 -I22 \\ I2 = \frac{sin6x}{144} - \frac{xcos6x}{24} - \frac{sin2x}{16} + \frac{xcos2x}{8} \\ I = I1 - I2 \\ I = \frac{sin4x}{64} - \frac{xcos4x}{16} - \frac{sin6x}{144} + \frac{xcos6x}{24} + \frac{sin2x}{16} - \frac{xcos2x}{8} + c \\ I = \frac{1}{8} | \frac{sin4x}{8} - \frac{xcos4x}{2} - \frac{sin6x}{18} + \frac{xcos6x}{3} + \frac{sin2x}{2} - xcos2x | + c$