1-e-tanx-t-1-t-2-dt-1-e-cotx-1-t-1-t-2-dt-

Question Number 101378 by Rohit@Thakur last updated on 02/Jul/20

\int_{\frac{1}{e}}^{t a n x} \frac{t}{1 + t^{2}} d t + \int_{\frac{1}{e}}^{c o t x} \frac{1}{t (1 + t^{2})} d t

Commented by Dwaipayan Shikari last updated on 02/Jul/20

\frac{1}{2} \int_{\frac{1}{e}}^{t a n x} \frac{2 t d t}{1 + t^{2}} = {[\frac{1}{2} l o g (1 + t^{2})]}_{\frac{1}{e}}^{t a n x} = \frac{1}{2} l o g ({s e c}^{2} x) - \frac{1}{2} l o g (\frac{1 + e^{2}}{e^{2}})

- \frac{1}{2} \int_{\frac{1}{e}}^{c o t x} \frac{\frac{- 2 d t}{t^{3}}}{1 + \frac{1}{t^{2}}} = {[- \frac{1}{2} l o g (1 + \frac{1}{t^{2}})]}_{\frac{1}{e}}^{c o t x} = - \frac{1}{2} l o g (\frac{{c o s e c}^{2} x}{{c o t}^{2} x}) + \frac{1}{2} l o g (1 + e^{2})

= - \frac{1}{2} l o g ({s e c}^{2} x) + \frac{1}{2} l o g (1 + e^{2})

S o \int_{\frac{1}{e}}^{t a n x} \frac{t}{1 + t^{2}} + \int_{\frac{1}{e}}^{c o t x} \frac{1}{t (1 + t^{2})} = \frac{1}{2} l o g (1 + e^{2}) - \frac{1}{2} l o g (\frac{1 + e^{2}}{e^{2}})

= \frac{1}{2} {l o g e}^{2} = 1 ★ L

Answered by 1549442205 last updated on 02/Jul/20

A = \int \frac{t}{1 + t^{2}} dt = \frac{1}{2} \int \frac{d (1 + t^{2})}{1 + t^{2}} = \frac{1}{2} \ln (1 + t^{2})

B = \int \frac{dt}{t (1 + t^{2})} . Putting t = \tan u \Rightarrow dt

\Rightarrow {\begin{cases} dt = (1 + t^{2}) du \\ \sin u = \sqrt{\frac{1}{1 + \cot^{2} u}} = \frac{1}{\sqrt{1 + \frac{1}{t^{2}}}} = \frac{t}{\sqrt{1 + t^{2}}} \end{cases}

\Rightarrow B = \int \frac{du}{\tan u} = \int \frac{\cos u}{\sin u} du = \int \frac{dsin u}{\sin u}

= \ln ∣ \sin u ∣= \ln ∣ \frac{t}{\sqrt{1 + t^{2}}} ∣ Hence,

\int_{\frac{1}{e}}^{t a n x} \frac{t}{1 + t^{2}} d t + \int_{\frac{1}{e}}^{c o t x} \frac{1}{t (1 + t^{2})} d t =

A (\tan x) - A (\frac{1}{e}) + B (cotx) - B (\frac{1}{e})

= \frac{1}{2} \ln \frac{1}{\cos^{2} x} - \frac{1}{2} \ln (1 + e^{- 2}) + \ln ∣ \cos x ∣

- \ln \frac{1}{\sqrt{1 + e^{2}}} = - \frac{1}{2} \ln (1 + e^{- 2}) + \ln \sqrt{1 + e^{2}}

= \ln \frac{\sqrt{1 + e^{2}}}{\sqrt{1 + e^{- 2}}} = lne = 1

Answered by mathmax by abdo last updated on 02/Jul/20

we have \int_{\frac{1}{e}}^{tanx} \frac{t}{1 + t^{2}} dt = {[\frac{1}{2} \ln (1 + t^{2})]}_{\frac{1}{e}}^{tanx} = \frac{1}{2} \ln (1 + \tan^{2} x) - \frac{1}{2} \ln (1 + e^{- 2})

= \frac{1}{2} \ln (\frac{1}{\cos^{2} x}) - \frac{1}{2} \ln (1 + e^{- 2}) = - \ln ∣ cosx ∣ - \frac{1}{2} \ln (1 + e^{- 2})

\int_{\frac{1}{e}}^{cotanx} \frac{dt}{t (1 + t^{2})} = \int_{\frac{1}{e}}^{\frac{1}{tanx}} (\frac{1}{t} - \frac{t}{1 + t^{2}}) dt = {[\ln ∣ t ∣]}_{\frac{1}{e}}^{\frac{1}{tanx}} - {[\frac{1}{2} \ln (1 + t^{2})]}_{\frac{1}{e}}^{\frac{1}{tanx}}

= - \ln ∣ tanx ∣ + 1 - \frac{1}{2} \ln (1 + \frac{1}{\tan^{2} x}) - \frac{1}{2} \ln (1 + e^{- 2})

= - \ln ∣ tanx ∣ - \frac{1}{2} \ln (1 + \tan^{2} x) + \ln ∣ tanx ∣ - \frac{1}{2} \ln (1 + e^{- 2}) + 1 \Rightarrow

\int_{\frac{1}{e}}^{tanx} \frac{tdt}{1 + t^{2}} + \int_{\frac{1}{e}}^{cotanx} \frac{dt}{t (1 + t^{2})} = \frac{1}{2} \ln (1 + \tan^{2} x) - \frac{1}{2} \ln (1 + e^{- 2})

- \frac{1}{2} \ln (1 + \tan^{2} x) - \frac{1}{2} \ln (1 + e^{- 2}) + 1 = 1