calculate-0-pi-2-cos-sin-cos-sin-d-

Question Number 37237 by abdo.msup.com last updated on 11/Jun/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{c o s θ . s i n θ}{c o s θ + s i n θ} d θ .

Commented by abdo.msup.com last updated on 27/Jul/18

c h a n g e m e n t t a n (\frac{θ}{2}) = t g i v e

I = \int_{0}^{1} \frac{\frac{2 t (1 - t^{2})}{{(1 + t^{2})}^{2}}}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

= \int_{0}^{1} \frac{2 t (1 - t^{2})}{{(1 + t^{2})}^{2} (1 - t^{2} + 2 t)} d t

= \int_{0}^{1} \frac{2 t (t^{2} - 1)}{{(1 + t^{2})}^{2} {(t - 1)}^{2}} d t

= \int_{0}^{1} \frac{2 t (t + 1)}{{(t^{2} + 1)}^{2} (t - 1)} d t

= \int_{0}^{1} \frac{2 t^{2} + 2 t}{(t - 1) {(t^{2} + 1)}^{2}} d t l e t d e c o m p o s e

F (t) = \frac{2 t^{2} + 2 t}{(t - 1) {(t^{2} + 1)}^{2}}

F (t) = \frac{a}{t - 1} + \frac{b t + c}{t^{2} + 1} + \frac{d t + e}{{(t^{2} + 1)}^{2}}

\dots . b e c o n t i n u e d \dots

Answered by math1967 last updated on 11/Jun/18

\frac{1}{2} \int \frac{2 s i n θ c o s θ}{c o s θ + s i n θ} d θ

\frac{1}{2} \int \frac{{(s i n θ + c o s θ)}^{2}}{c o s θ + s i n θ} d θ - \frac{1}{2} \int \frac{d θ}{c o s θ + s i n θ}

- \frac{1}{2} c o s θ + \frac{1}{2} s i n θ - \frac{1}{2 \sqrt{2}} \int \frac{d θ}{s i n \frac{π}{4} c o s θ + c o s \frac{π}{4} s i n θ}

\frac{1}{2} (s i n θ - c o s θ) - \frac{1}{2 \sqrt{2}} \int c o s e c (\frac{π}{4} + θ) d θ

\frac{1}{2} (s i n θ - c o s θ) - \frac{1}{2 \sqrt{2}} l n [c o s e c (\frac{π}{4} + θ) - c o t (\frac{π}{4} + θ)]

∴ \underset{0}{\int^{\frac{π}{2}}} \frac{s i n θ c o s θ d θ}{s i n θ + c o s θ} = {\frac{1}{2} \times 1 - \frac{1}{2 \sqrt{2}} l n [\sqrt{2} + 1]}

- {- \frac{1}{2} - \frac{1}{2 \sqrt{2}} l n [[\sqrt{2} - 1]} = 1 + \frac{1}{2 \sqrt{2}} l n \frac{\sqrt{2} - 1}{\sqrt{2} + 1}

Commented by math1967 last updated on 11/Jun/18

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