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Question Number 98426 by mathmax by abdo last updated on 13/Jun/20

let f (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{x + tant} calculate f (x)

2) explicit g (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{{(x + tant)}^{2}}

3) find the value of integrals \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{2 + tant} and \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{{(2 + tant)}^{2}}

Answered by maths mind last updated on 14/Jun/20

f (x) = \int_{1}^{\sqrt{3}} \frac{d y}{(x + y) (1 + y^{2})}

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

= \frac{1}{1 + x^{2}} \int_{1}^{\sqrt{3}} {\frac{x}{1 + y^{2}} - \frac{y}{1 + y^{2}} + \frac{1}{x + y}} d y

= \frac{x}{1 + x^{2}} [\frac{π}{12}] - \frac{l n (2)}{2 (1 + x^{2})} + \frac{l n (\frac{x + \sqrt{3}}{x + 1})}{(1 + x^{2})}

g (x) = - f^{'} (x)

= - {\frac{1 - x^{2}}{{(1 + x^{2})}^{2}} \frac{π}{12} + \frac{x l n (2)}{{(1 + x^{2})}^{2}} + \frac{1}{(1 + x^{2})} [\frac{1}{x + \sqrt{3}} - \frac{1}{x + 1}] - \frac{2 x l n (\frac{x + \sqrt{3}}{x + 1})}{{(1 + x^{2})}^{2}}}

3 \int \frac{d x}{2 + t g (x)} = f (2) = \frac{π}{30} - \frac{l n (2)}{10} + \frac{l n (\frac{2 + \sqrt{3}}{3})}{5}

\int \frac{d x}{{(2 + t g (x))}^{2}} = - {\frac{- π}{100} . + \frac{2 l n (2)}{25} + \frac{1}{5} (\frac{1}{2 + \sqrt{3}} - \frac{1}{3}) - \frac{4 l n (\frac{2 + \sqrt{3}}{3})}{25}}

Commented by abdomathmax last updated on 14/Jun/20

thanks sir .

Commented by maths mind last updated on 14/Jun/20

w i t h e p l e a s e r

Answered by abdomathmax last updated on 14/Jun/20

1) f (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{x + tant} changement tant = u give

f (x) = \int_{1}^{\sqrt{3}} \frac{du}{(1 + u^{2}) (x + u)} let decompose

F (u) = \frac{1}{(u + x) (u^{2} + 1)} \Rightarrow F (u) = \frac{a}{u + x} + \frac{bu + c}{u^{2} + 1}

a = \frac{1}{x^{2} + 1} \lim_{u \to + \infty} uF (u) = 0 = a + b \Rightarrow b = - \frac{1}{x^{2} + 1}

F (0) = \frac{1}{x} = \frac{a}{x} + c \Rightarrow 1 = a + xc \Rightarrow xc = 1 - a

= 1 - \frac{1}{x^{2} + 1} = \frac{x^{2}}{1 + x^{2}} \Rightarrow c = \frac{x}{1 + x^{2}} \Rightarrow

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

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

- \frac{1}{x^{2} + 1} \int_{1}^{\sqrt{3}} \frac{u - x}{u^{2} + 1} du

= \frac{1}{x^{2} + 1} {[\ln ∣ u + x ∣]}_{1}^{\sqrt{3}} - \frac{1}{2 (x^{2} + 1)} {[\ln (u^{2} + 1)]}_{1}^{\sqrt{3}}

+ \frac{x}{x^{2} + 1} {[arctanu]}_{1}^{\sqrt{3}}

= \frac{1}{x^{2} + 1} {\ln ∣ \sqrt{3} + x ∣ - \ln ∣ 1 + x ∣}

- \frac{2 \ln (2)}{2 (x^{2} + 1)} + \frac{x}{x^{2} + 1} \times \frac{π}{12}

= \frac{1}{x^{2} + 1} \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣ - \frac{\ln 2}{x^{2} + 1} + \frac{π x}{12 (x^{2} + 1)}

Commented by abdomathmax last updated on 14/Jun/20

2) we have f^{^{'}} (x) = - \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{{(3 + tant)}^{2}} = - g (x)

\Rightarrow g (x) = - f^{^{'}} (x) we have

f (x) = \frac{1}{x^{2} + 1} {π x - \ln 2 + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣} \Rightarrow

f^{^{'}} (x) = - \frac{2 x}{{(x^{2} + 1)}^{2}} {π x - \ln (2) + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣}

+ \frac{1}{x^{2} + 1} {π + \frac{{(\frac{x + \sqrt{3}}{x + 1})}^{^{'}}}{\frac{x + \sqrt{3}}{x + 1}}}

= \frac{- 2 x}{{(x^{2} + 1)}^{2}} {π x - \ln 2 + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣}

+ \frac{1}{x^{2} + 1} {π + \frac{(x + 1)}{x + \sqrt{3}} \times \frac{x + 1 - (x + \sqrt{3})}{{(x + 1)}^{2}}}

= \frac{- 2 x}{{(x^{2} + 1)}^{2}} {π x - \ln 2 + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣}

+ \frac{1}{x^{2} + 1} {π + \frac{1 - \sqrt{3}}{(x + \sqrt{3}) (x + 1)}}

Commented by mathmax by abdo last updated on 14/Jun/20

\Rightarrow g (x) = \frac{2 x}{{(x^{2} + 1)}^{2}} {π x - \ln 2 + \ln ∣ \frac{x + \sqrt{3}}{x + 1} ∣} - \frac{1}{x^{2} + 1} {π + \frac{1 - \sqrt{3}}{(x + \sqrt{3}) (x + 1)}}

Commented by mathmax by abdo last updated on 14/Jun/20

3) \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{dt}{2 + tant} = f (2) = \frac{1}{5} \ln (\frac{2 + \sqrt{3}}{3}) - \frac{\ln 2}{5} + \frac{2 π}{12 (5)}

= \frac{1}{5} \ln (\frac{2 + \sqrt{3}}{3}) - \frac{\ln (2)}{5} + \frac{π}{30}