advanced-calculus-calculate-I-0-pi-x-1-sin-x-cos-x-dx-

Question Number 123023 by mnjuly1970 last updated on 21/Nov/20

\dots a d v a n c e d c a l c u l u s \dots

c a l c u l a t e :::

I : \overset{? ? ?}{=} \int_{0}^{π} \frac{x}{1 - s i n (x) c o s (x)} d x

\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots . .

Answered by mnjuly1970 last updated on 24/Nov/20

e v a l u a t i o n : I \overset{?}{=} \int_{0}^{π} \frac{x}{1 - s i n (x) c o s (x)} d x

s o l u t i o n :

I = [\int_{0}^{\frac{π}{2}} \frac{x}{1 - s i n (x) c o s (x)} d x = I_{1}] + [\int_{\frac{π}{2}}^{π} \frac{x}{1 - x s i n (x) c o s (x)} d x = I_{2}]

I_{1} = \int_{0}^{\frac{π}{2}} \frac{2 x}{2 - s i n (2 x)} d x = π \int_{0}^{\frac{π}{2}} \frac{1}{2 - s i n (2 x)} d x - 2 \int_{0}^{\frac{π}{2}} \frac{x}{2 - s i n (2 x)} d x

{2 I}_{1} = π \int_{0}^{\frac{π}{2}} \frac{d x}{2 - \frac{2 t a n (x)}{1 + {t a n}^{2} (x)}} = π \int_{0}^{\frac{π}{2}} \frac{1 + {t a n}^{2} (x)}{2 {t a n}^{2} (x) - 2 t a n (x) + 2} d x

Missing \left or extra \right

= \frac{π}{\sqrt{3}} [\frac{π}{2} + \frac{π}{6}] = \frac{2 π^{2}}{3 \sqrt{3}} \Rightarrow I_{1} = \frac{π^{2}}{3 \sqrt{3}} ✓

I_{2} = \int_{\frac{π}{2}}^{π} \frac{x}{1 - s i n (x) c o s (x)} d x

= \frac{3 π}{2} \int_{\frac{π}{2}}^{π} \frac{2}{2 - s i n (2 x)} d x - I_{2}

{2 I}_{2} \underset{I_{1}}{\overset{s i m i l a r l y}{=}} (\frac{3 π}{2} (\frac{2}{\sqrt{3}}) [{t a n}^{- 1} {(\frac{2}{\sqrt{3}} (t - \frac{1}{2})]}_{- \infty}^{0}

{2 I}_{2} = \frac{3 π}{\sqrt{3}} [\frac{- π}{6} + \frac{π}{2}] = \frac{3 π^{2}}{3 \sqrt{3}} \Rightarrow I_{2} = \frac{π^{2}}{2 \sqrt{3}}

∴ I = I_{1} + I_{2} = \frac{π^{2}}{\sqrt{3}} (\frac{1}{2} + \frac{1}{3}) = \frac{5 π^{2}}{6 \sqrt{3}} ✓ ✓

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