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Question Number 64448 by mathmax by abdo last updated on 18/Jul/19

c a l c u l a t e {l i m}_{x \to π} \int_{\frac{π}{2}}^{x} \frac{d x}{1 + s i n x - c o s x}

Commented by mathmax by abdo last updated on 19/Jul/19

l e t A (x) = \int_{\frac{π}{2}}^{x} \frac{d t}{1 + s i n t - c o s t} c h a n g e m e n t t a n (\frac{t}{2}) = u v i v e

A (x) = \int_{1}^{t a n (\frac{x}{2})} \frac{2 d u}{(1 + u^{2}) (1 + \frac{2 u}{1 + u^{2}} - \frac{1 - u^{2}}{1 + u^{2}})}

= \int_{1}^{t a n (\frac{x}{2})} \frac{2 d u}{1 + u^{2} + 2 u - 1 + u^{2}} = \int_{1}^{t a n (\frac{x}{2})} \frac{2 d u}{2 u^{2} + 2 u} = \int_{1}^{t a n (\frac{x}{2})} \frac{d u}{u (u + 1)}

= \int_{1}^{t a n (\frac{x}{2})} {\frac{1}{u} - \frac{1}{u + 1}} d u = {[l n ∣ \frac{u}{u + 1} ∣]}_{1}^{t a n (\frac{x}{2})} = l n ∣ \frac{t a n (\frac{x}{2})}{t a n (\frac{x}{2}) + 1} ∣ - l n (\frac{1}{2})

\Rightarrow {l i m}_{x \to π} A (x) = 0 + l n (2) = l n (2) .

Answered by Tanmay chaudhury last updated on 18/Jul/19

\int \frac{d x}{2 s i n \frac{x}{2} c o s \frac{x}{2} + 2 {s i n}^{2} \frac{x}{2}}

\int \frac{{s e c}^{2} \frac{x}{2} d x}{2 t a n \frac{x}{2} + 2 {t a n}^{2} \frac{x}{2}} d x

\int \frac{d (t a n \frac{x}{2})}{t a n \frac{x}{2} (1 + t a n \frac{x}{2})}

\int \frac{d (t a n \frac{x}{2})}{t a n \frac{x}{2}} - \int \frac{d (t a n \frac{x}{2})}{1 + t a n \frac{x}{2}}

∣ l n (\frac{t a n \frac{x}{2}}{1 + t a n \frac{x}{2}}) ∣_{\frac{π}{2}}^{π}

l n 1 - l n (\frac{1}{1 + 1})

= l n 2