I-0-1-1-x-x-2-x-2-1-2-dx-find-tan-I-sec-I-

Question Number 96679 by M±th+et+s last updated on 03/Jun/20

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

f i n d t a n (I) + s e c (I)

Answered by mathmax by abdo last updated on 03/Jun/20

I = \int_{0}^{1} \frac{1 - x}{x^{2} + {(x^{2} + 1)}^{2}} dx \Rightarrow I = \int_{0}^{1} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} - \int_{0}^{1} \frac{xdx}{x^{2} + {(x^{2} + 1)}^{2}} = H - K

\int_{0}^{1} \frac{xdx}{x^{2} + {(x^{2} + 1)}^{2}} = \int_{0}^{1} \frac{xdx}{x^{2} + 1 + {(x^{2} + 1)}^{2} - 1} =_{x^{2} + 1 = u} \int_{1}^{2} \frac{du}{2 (u + u^{2} - 1)}

= \frac{1}{2} \int_{1}^{2} \frac{du}{u^{2} + u - 1}

u^{2} + u - 1 = 0 \to Δ = 1 + 4 = 5 \Rightarrow u_{1} = \frac{- 1 + \sqrt{5}}{2} and u_{2} = \frac{- 1 - \sqrt{5}}{2}

\int_{1}^{2} \frac{du}{u^{2} + u - 1} = \frac{1}{\sqrt{5}} \int_{1}^{2} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) du = \frac{1}{\sqrt{5}} {[\ln ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{1}^{2}

= \frac{1}{\sqrt{5}} {\ln ∣ \frac{2 - u_{1}}{2 - u_{2}} ∣ - \ln ∣ \frac{1 - u_{1}}{1 - u_{2}} ∣} = \frac{1}{\sqrt{5}} {\ln ∣ \frac{2 - \frac{- 1 + \sqrt{5}}{2}}{2 + \frac{1 + \sqrt{5}}{2}} ∣ - \ln ∣ \frac{1 - \frac{- 1 + \sqrt{5}}{2}}{1 + \frac{1 + \sqrt{5}}{2}} ∣} \Rightarrow

\int_{0}^{1} \frac{xdx}{x^{2} + {(x^{2} + 1)}^{2}} dx = \frac{1}{2 \sqrt{5}} {\ln ∣ \frac{5 - \sqrt{5}}{5 + \sqrt{5}} ∣ - \ln ∣ \frac{3 - \sqrt{5}}{3 + \sqrt{5}} ∣} = K

H = \int_{0}^{1} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} ? we have \int_{0}^{\infty} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}}

= \int_{0}^{1} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} + \int_{1}^{+ \infty} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} (\to x = \frac{1}{t})

= \int_{0}^{1} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} - \int_{0}^{1} \frac{dt}{t^{2} (\frac{1}{t^{2}} + {(\frac{1}{t^{2}} + 1)}^{2})}

= \int_{0}^{1} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} - \int_{0}^{1} \frac{dt}{1 + \frac{{(1 + t^{2})}^{2}}{t^{2}}} = \int_{0}^{1} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} - \int_{0}^{1} \frac{t^{2} dt}{t^{2} + {(1 + t^{2})}^{2}} \Rightarrow

\int_{0}^{1} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} = \int_{0}^{\infty} \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} + \int_{0}^{1} \frac{x^{2} dx}{x^{2} + {(1 + x^{2})}^{2}} \dots . be continued \dots

Commented by 1549442205 last updated on 04/Jun/20

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

f i n d t a n (I) + s e c (I)

J = \int \frac{dx}{x^{2} + {(x^{2} + 1)}^{2}} \underset{x = tant}{=} \int \frac{(1 + \tan^{2} t) dt}{\tan^{2} t + {(\tan^{2} t + 1)}^{2}}

= \int \frac{dt}{1 + \tan^{2} t + 1 - \frac{1}{\tan^{2} t + 1}} = \int \frac{dt}{\frac{1}{\cos^{2} t} + 1 - \cos^{2} t}

= \int \frac{\cos^{2} tdt}{1 + {(sint . cost)}^{2}} = \int \frac{2 (1 + \cos 2 t) dt}{{(2 sint . cost)}^{2} + 4} =

\int \frac{2 dt}{\sin^{2} 2 t + 4} + \int \frac{dsin 2 t}{\sin^{2} 2 t + 4} = M + \frac{\sin 2 t}{2} . \arcsin (\frac{\sin 2 t}{2}) =

M + \frac{x}{1 + x^{2}} . \arcsin \frac{x}{1 + x^{2}}

M = \int \frac{2 dt}{\sin^{2} 2 t + 4} = 2 \int \frac{dt}{\frac{1 - \cos 4 t}{2} + 4} = - 4 \int \frac{dt}{\cos 4 t - 9}

\underset{u = \tan 2 t}{=} - 2 \int \frac{du}{(1 + u^{2}) (\frac{1 - u^{2}}{1 + u^{2}} - 9)}

= \int \frac{- 2 du}{1 - u^{2} - 9 (1 + u^{2})} = \int \frac{du}{{5 u}^{2} + 4} = \frac{1}{5} \int \frac{du}{u^{2} + {(\frac{2}{\sqrt{5}})}^{2}}

= \frac{1}{5} [\frac{\sqrt{5} u}{2} . \arcsin (\frac{\sqrt{5} u}{2})] = \frac{1}{5} . \frac{\sqrt{5} x}{1 - x^{2}} . \arcsin \frac{\sqrt{5} x}{1 - x^{2}}

Hence, J = \frac{x}{1 + x^{2}} \arcsin \frac{x}{1 + x^{2}} + \frac{1}{5} . \frac{\sqrt{5} x}{1 - x^{2}} . \arcsin \frac{\sqrt{5} x}{1 - x^{2}}

K = \int \frac{xdx}{x^{2} + {(x^{2} + 1)}^{2}} \underset{u = x^{2} + 1}{=} \int \frac{du}{u^{2} + u - 1} = \int \frac{du}{{(u + \frac{1}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2}}

= \frac{1}{\sqrt{5}} \ln ∣ \frac{u + \frac{1}{2} - \frac{\sqrt{5}}{2}}{u + \frac{1}{2} + \frac{\sqrt{5}}{2}} ∣= \frac{1}{\sqrt{5}} \ln ∣ \frac{{2 x}^{2} + 3 - \sqrt{5}}{{2 x}^{2} + 3 + \sqrt{5}} ∣

Therefore, F (x) = J + N = \frac{x}{1 + x^{2}} . \arcsin \frac{x}{1 + x^{2}} + \frac{1}{5} . \frac{\sqrt{5} x}{1 - x^{2}} . \arcsin \frac{\sqrt{5} x}{1 - x^{2}} + \frac{1}{\sqrt{5}} \ln ∣ \frac{{2 x}^{2} + 3 - \sqrt{5}}{{2 x}^{2} + 3 + \sqrt{5}} ∣ + C

F (1) = \frac{1}{2} . \frac{π}{6} + 0 + \frac{1}{\sqrt{5}} . \ln \frac{\sqrt{5} - 1}{\sqrt{5} + 1} + C

F (0) = \frac{1}{\sqrt{5}} . \ln \frac{3 - \sqrt{5}}{3 + \sqrt{5}} + C \Rightarrow I = F (1) - F (0) = \frac{π}{12} + \frac{1}{\sqrt{5}} \ln \frac{\sqrt{5} - 1}{\sqrt{5} + 1} - \frac{1}{\sqrt{5}} \ln \frac{3 - \sqrt{5}}{3 + \sqrt{5}}

= \frac{π}{12} + \frac{2}{\sqrt{5}} \ln \frac{\sqrt{5} + 1}{2} \approx 0.6922083288

So, \tan (I) + \sec (I) \approx 2.128029314

Commented by M±th+et+s last updated on 04/Jun/20

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