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Question Number 72888 by mathmax by abdo last updated on 04/Nov/19

l e t f (x) = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{t a n t}{2 + x c o s t} d t w i t h x r e a l

1) d e t e r m i n e a e x p l i c i t f o r m f o r f (x)

2) d e t e r m i n e a l s o g (x) = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{t a n t}{{(2 + x c o s t)}^{2}} d x

3) f i n d t h e v a l u e o f \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{t a n t}{(2 + 3 c o s t)} d t a n d \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{t a n t}{{(2 + 3 c o s t)}^{2}} d t

Commented by mathmax by abdo last updated on 05/Nov/19

w e h a v e f (x) = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{s i n t}{c o s t (2 + x c o s t)} d t c h a n g e n t c o s t = u g i v e

f (x) = \int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{2}}{2}} \frac{- d u}{u (2 + x u)} = - \int_{\frac{\sqrt{3}}{3}}^{\frac{\sqrt{2}}{2}} \frac{d u}{u (x u + 2)} l e t d e c o m p o s e

F (u) = \frac{1}{u (x u + 2)} \Rightarrow F (u) = \frac{a}{u} + \frac{b}{x u + 2}

a = \frac{1}{2} a n d b = {l i m}_{u \to - \frac{2}{x}} (x u + 2) F (u) = \frac{1}{- \frac{2}{x}} = - \frac{x}{2} \Rightarrow

F (u) = \frac{1}{2 u} - \frac{x}{2 (x u + 2)} \Rightarrow f (x) = - \frac{1}{2} \int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{2}}{2}} (\frac{1}{u} - \frac{x}{x u + 2}) d u

= - \frac{1}{2} {[l n ∣ \frac{u}{x u + 2} ∣]}_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{2}}{2}} = - \frac{1}{2} {l n ∣ \frac{\frac{\sqrt{2}}{2}}{x \frac{\sqrt{2}}{2} + 2} ∣ - l n ∣ \frac{\frac{\sqrt{3}}{2}}{x \frac{\sqrt{3}}{2} + 2} ∣}

= - \frac{1}{2} {l n ∣ \frac{\sqrt{2}}{4 + x \sqrt{2}} ∣ - l n ∣ \frac{\sqrt{3}}{4 + x \sqrt{3}} ∣}

= - \frac{1}{2} {\frac{1}{2} l n (2) - l n ∣ 4 + x \sqrt{2} ∣ - \frac{1}{2} l n (3) + l n ∣ 4 + x \sqrt{3} ∣}

= - \frac{1}{4} (l n (2) - l n (3)) + \frac{1}{2} l n ∣ \frac{4 + x \sqrt{2}}{4 + x \sqrt{3}} ∣ \Rightarrow

f (x) = \frac{1}{4} l n (\frac{3}{2}) + \frac{1}{2} l n ∣ \frac{4 + x \sqrt{2}}{4 + x \sqrt{3}} ∣

Commented by mathmax by abdo last updated on 05/Nov/19

2) w e h a v e f^{^{'}} (x) = - \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{c o s t \times t a n t}{{(2 + x c o s t)}^{2}} d t = - \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{s i n t}{{(2 + x c o s t)}^{2}} d t

= - g (x) \Rightarrow g (x) = - f^{^{'}} (x) b u t

f^{^{'}} (x) = \frac{1}{2} \frac{{(\frac{4 + x \sqrt{2}}{4 + x \sqrt{3}})}^{^{'}}}{\frac{4 + x \sqrt{2}}{4 + x \sqrt{3}}} = \frac{4 + x \sqrt{3}}{2 (4 + x \sqrt{2})} \times \frac{\sqrt{2} (4 + x \sqrt{3}) - \sqrt{3} (4 + x \sqrt{2})}{{(4 + x \sqrt{3})}^{2}}

= \frac{1}{2 (4 + x \sqrt{2}) (4 + x \sqrt{3})} \times 4 (\sqrt{2} - \sqrt{3}) = \frac{2 \sqrt{2} - 2 \sqrt{3}}{(4 + x \sqrt{2}) (4 + x \sqrt{3})} \Rightarrow

g (x) = \frac{2 \sqrt{3} - 2 \sqrt{2}}{(4 + x \sqrt{2}) (4 + x \sqrt{3})}

Commented by mathmax by abdo last updated on 05/Nov/19

3) \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{t a n t}{(2 + 3 c o s t)} d t = f (3) = \frac{1}{4} l n (\frac{3}{2}) + \frac{1}{2} l n ∣ \frac{4 + 3 \sqrt{2}}{4 + 3 \sqrt{3}} ∣

Commented by mathmax by abdo last updated on 05/Nov/19

\int_{\frac{π}{6}}^{\frac{π}{4}} \frac{s i n t}{{(2 + 3 c o s t)}^{2}} d t = g (3) = \frac{2 \sqrt{3} - 2 \sqrt{2}}{(4 + 3 \sqrt{2}) (4 + 3 \sqrt{3})}

Commented by mathmax by abdo last updated on 05/Nov/19

s o r r y g (x) = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{s i n t}{{(2 + x c o s t)}^{2}} d t

Answered by mind is power last updated on 04/Nov/19

f (x) = \int \frac{\sin (t) dt}{\cos (t) (2 + xcos (t))}

u = \cos (t)

f (x) = \int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{2}}{2}} \frac{- du}{u (2 + xu)} = \int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}} \frac{du}{2 u} - \frac{xdu}{2 (2 + xu)} = \frac{1}{2} \ln (\frac{\sqrt{3}}{\sqrt{2}}) - \frac{1}{2} \ln (\frac{2 + \frac{x \sqrt{3}}{2}}{2 + x \frac{\sqrt{2}}{2}})

f^{'} (x) = \int - \frac{\tan (t) \cos (t)}{{(2 + xcos (t))}^{2}} dt

= - \int \frac{\tan (t) (\cos (t) + \frac{2}{x} - \frac{2}{x})}{{(2 + xcos (t))}^{2}} = - \frac{1}{x} \int \frac{\tan (t)}{(2 + xcos (t))} + \frac{2}{x} \int \frac{\tan (t)}{{(2 + xcos (t))}^{2}}

\Rightarrow \frac{{xf}^{'} (x)}{2} + \frac{f (x)}{2} = g (x)

g (x) = \frac{x}{2} {- \frac{\sqrt{3}}{4 + x \sqrt{2}} + \frac{\sqrt{2}}{4 + x \sqrt{2}}} + \frac{1}{2} {\ln (\frac{\sqrt{3}}{\sqrt{2}}) - \frac{1}{2} \ln (\frac{4 + x \sqrt{3}}{4 + x \sqrt{2}})}

\int_{\frac{π}{6}}^{\frac{π}{4}} \frac{\tan (t)}{(2 + 3 \cos (t))} dt = f (3) = \frac{1}{2} \ln (\frac{\sqrt{3}}{\sqrt{2}}) - \frac{1}{2} \ln (\frac{4 + 3 \sqrt{3}}{4 + 3 \sqrt{2}})

\int_{\frac{π}{6}}^{\frac{π}{4}} \frac{(\tan (t))}{{(2 + 3 \cos (t))}^{2}} dt = g (3) = \frac{3}{2} (\frac{\sqrt{2} - \sqrt{3}}{4 + 3 \sqrt{2}}) + \frac{\ln (\frac{\sqrt{3}}{\sqrt{2}})}{2} - \frac{1}{4} \ln (\frac{4 + 3 \sqrt{3}}{4 + 3 \sqrt{3}})

Commented by turbo msup by abdo last updated on 04/Nov/19

t h s n k y o u s i r .

Commented by mind is power last updated on 04/Nov/19

y, re welcom