find-f-x-pi-4-pi-3-cosxdx-2cos-2-x-sin-2-x-1-

Question Number 42801 by maxmathsup by imad last updated on 02/Sep/18

f i n d f (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{c o s x d x}{2 {c o s}^{2} x + {s i n}^{2} x + 1}

Commented by maxmathsup by imad last updated on 04/Sep/18

l e t I = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{c o s x d x}{2 (1 - {s i n}^{2} x) + {s i n}^{2} x + 1} c h a n g e m e n t s i n x = t g i v e

I = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{2 (1 - t^{2}) + t^{2} + 1} = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{3 - t^{2}} = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{(\sqrt{3} - t) (\sqrt{3} + t)} d t

= \frac{1}{2 \sqrt{3}} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} {\frac{1}{\sqrt{3} - t} + \frac{1}{\sqrt{3} + t}} d t = \frac{1}{2 \sqrt{3}} {[l n ∣ \frac{\sqrt{3} + t}{\sqrt{3} - t} ∣]}_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} = \frac{1}{2 \sqrt{3}} {l n ∣ \frac{\sqrt{3} + \frac{\sqrt{3}}{2}}{\sqrt{3} - \frac{\sqrt{3}}{2}} ∣

- l n ∣ \frac{\sqrt{3} + \frac{1}{\sqrt{2}}}{\sqrt{3} - \frac{1}{\sqrt{2}}} ∣} = \frac{1}{2 \sqrt{3}} {l n (\frac{3 \sqrt{3}}{\sqrt{3}}) - l n (\frac{\sqrt{6} + 1}{\sqrt{6} - 1})} = \frac{1}{2 \sqrt{3}} {l n (3) - l n (\frac{1 + \sqrt{6}}{- 1 + \sqrt{6}})} .

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18

\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{2 (1 - t^{2}) + t^{2} + 1} t = s i n x d t = c o s x d x

\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{3 - t^{2}} =∣ \frac{1}{2 \sqrt{3}} l n (\frac{t + \sqrt{3}}{t - \sqrt{3}}) ∣_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}}

\frac{1}{2 \sqrt{3}} {l n ∣ (\frac{\frac{\sqrt{3}}{2} + \sqrt{3}}{\frac{\sqrt{3}}{2} - \sqrt{3}}) ∣ - l n ∣ (\frac{\frac{1}{\sqrt{2}} + \sqrt{3}}{\frac{1}{\sqrt{2}} - \sqrt{3}}) ∣}

= \frac{1}{2 \sqrt{3}} {l n ∣ (\frac{\frac{3}{2}}{- \frac{1}{2}}) ∣ - l n ∣ (\frac{1 + \sqrt{6}}{1 - \sqrt{6}}) ∣}

= \frac{1}{2 \sqrt{3}} {l n ∣ - 3 ∣ - l n ∣ (\frac{1 + \sqrt{6}}{1 - \sqrt{6}})}

u s e f o r m u l a \int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} \int \frac{a + x + a - x}{(a + x) (a - x)} d x

\frac{1}{2 a} [\int \frac{d x}{a - x} + \int \frac{d x}{a + x}]

\frac{1}{2 a} [\int \frac{d x}{x + a} - \int \frac{d x}{x - a}] = \frac{1}{2 a} l n (\frac{x + a}{x - a})