Thr-value-of-0-pi-2-cosec-x-pi-3-cosec-x-pi-6-dx-is-

Question Number 75924 by benjo last updated on 21/Dec/19

Thr value of

\underset{0}{\int^{π / 2}} cosec (x - \frac{π}{3}) cosec (x - \frac{π}{6}) d x is

Commented by mathmax by abdo last updated on 21/Dec/19

I = \int_{0}^{\frac{π}{2}} \frac{d x}{s i n (x - \frac{π}{3}) s i n (x - \frac{π}{6})} b u t w e h a v e

s i n (x - \frac{π}{3}) s i n (x - \frac{π}{6}) = c o s (\frac{π}{2} - x + \frac{π}{3}) c o s (\frac{π}{2} - x + \frac{π}{6})

= c o s (\frac{5 π}{6} - x) c o s (\frac{2 π}{3} - x)

= \frac{1}{2} {c o s (\frac{5 π}{6} + \frac{2 π}{3} - 2 x) + c o s (\frac{5 π}{6} - x - \frac{2 π}{3} + x)}

= \frac{1}{2} {c o s (\frac{3 π}{2} - 2 x) + c o s (\frac{π}{6})} = \frac{1}{2} {c o s (- \frac{π}{2} - 2 x) + \frac{\sqrt{3}}{2}}

= \frac{1}{2} {\frac{\sqrt{3}}{2} - s i n (2 x)} \Rightarrow I = 2 \int_{0}^{\frac{π}{2}} \frac{d x}{\frac{\sqrt{3}}{2} - s i n (2 x)}

= 4 \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{3} - 2 s i n (2 x)} =_{2 x = t} 4 \int_{0}^{π} \frac{d t}{2 (\sqrt{3} - 2 s i n t)}

= 2 \int_{0}^{π} \frac{d t}{\sqrt{3} - 2 s i n (t)} =_{t a n (\frac{t}{2}) = u} 2 \int_{0}^{+ \infty} \frac{1}{\sqrt{3} - 2 \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= 4 \int_{0}^{\infty} \frac{d u}{\sqrt{3} + \sqrt{3} u^{2} - 4 u} = 4 \int_{0}^{\infty} \frac{d u}{\sqrt{3} u^{2} - 4 u + \sqrt{3}}

Δ^{^{'}} = {(- 2)}^{2} - 3 = 1 \Rightarrow u_{1} = \frac{2 + 1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}

u_{2} = \frac{2 - 1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow I = 4 \int_{0}^{\infty} \frac{d u}{\sqrt{3} (u - \sqrt{3}) (u - \frac{1}{\sqrt{3}})}

= \frac{4}{\sqrt{3}} \times \frac{1}{\sqrt{3} - \frac{1}{\sqrt{3}}} \int_{0}^{\infty} (\frac{1}{u - \sqrt{3}} - \frac{1}{u - \frac{1}{\sqrt{3}}}) d u

= 2 l n ∣ \frac{u - \sqrt{3}}{u - \frac{1}{\sqrt{3}}} ∣ + c

= 2 l n ∣ \frac{t a n (x) - \sqrt{3}}{t a n (x) - \frac{1}{\sqrt{3}}} ∣ + c

Commented by benjo last updated on 22/Dec/19

thanks sir

Commented by abdomathmax last updated on 23/Dec/19

y o u a r e w e l c o m e .