∫_(1/e) ^(tanx) (t/(1+t^2 ))dt + ∫_(1/e) ^(cotx) (1/(t(1+t^2 )))dt


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
	$A=∫(t/(1+t^2 ))dt=(1/2)∫((d(1+t^2 ))/(1+t^2 ))=(1/2)ln(1+t^2 ) B=∫(dt/(t(1+t^2 ))).Putting t=tan u⇒dt ⇒ { ((dt=(1+t^2 )du)),((sin u=(√(1/(1+cot^2 u)))=(1/(√(1+(1/t^2 ))))=(t/(√(1+t^2 ))))) :} ⇒B=∫(du/(tan u))=∫((cos u)/(sin u))du=∫((dsin u)/(sin u)) =ln∣sin u∣=ln∣(t/(√(1+t^2 )))∣Hence, ∫_(1/e) ^(tanx) (t/(1+t^2 ))dt + ∫_(1/e) ^(cotx) (1/(t(1+t^2 )))dt= A(tan x)−A((1/e))+B(cotx)−B((1/e)) =(1/2)ln(1/(cos^2 x))−(1/2)ln(1+e^(−2) )+ln∣cos x∣ −ln(1/(√(1+e^2 )))=−(1/2)ln(1+e^(−2) )+ln(√(1+e^2 )) =ln((√(1+e^2 ))/(√(1+e^(−2) )))=lne=1$
	$A = \int \frac{t}{1 + t^{2}} dt = \frac{1}{2} \int \frac{d (1 + t^{2})}{1 + t^{2}} = \frac{1}{2} \ln (1 + t^{2})$ $B = \int \frac{dt}{t (1 + t^{2})} . Putting t = \tan u \Rightarrow dt$ $\Rightarrow {\begin{cases} dt = (1 + t^{2}) du \\ \sin u = \sqrt{\frac{1}{1 + \cot^{2} u}} = \frac{1}{\sqrt{1 + \frac{1}{t^{2}}}} = \frac{t}{\sqrt{1 + t^{2}}} \end{cases}$ $\Rightarrow B = \int \frac{du}{\tan u} = \int \frac{\cos u}{\sin u} du = \int \frac{dsin u}{\sin u}$ $= \ln ∣ \sin u ∣= \ln ∣ \frac{t}{\sqrt{1 + t^{2}}} ∣ Hence,$ $\int_{\frac{1}{e}}^{t a n x} \frac{t}{1 + t^{2}} d t + \int_{\frac{1}{e}}^{c o t x} \frac{1}{t (1 + t^{2})} d t =$ $A (\tan x) - A (\frac{1}{e}) + B (cotx) - B (\frac{1}{e})$ $= \frac{1}{2} \ln \frac{1}{\cos^{2} x} - \frac{1}{2} \ln (1 + e^{- 2}) + \ln ∣ \cos x ∣$ $- \ln \frac{1}{\sqrt{1 + e^{2}}} = - \frac{1}{2} \ln (1 + e^{- 2}) + \ln \sqrt{1 + e^{2}}$ $= \ln \frac{\sqrt{1 + e^{2}}}{\sqrt{1 + e^{- 2}}} = lne = 1$
	Answered by mathmax by abdo last updated on 02/Jul/20

	$we have \int_{\frac{1}{e}}^{tanx} \frac{t}{1 + t^{2}} dt = {[\frac{1}{2} \ln (1 + t^{2})]}_{\frac{1}{e}}^{tanx} = \frac{1}{2} \ln (1 + \tan^{2} x) - \frac{1}{2} \ln (1 + e^{- 2})$ $= \frac{1}{2} \ln (\frac{1}{\cos^{2} x}) - \frac{1}{2} \ln (1 + e^{- 2}) = - \ln ∣ cosx ∣ - \frac{1}{2} \ln (1 + e^{- 2})$ $\int_{\frac{1}{e}}^{cotanx} \frac{dt}{t (1 + t^{2})} = \int_{\frac{1}{e}}^{\frac{1}{tanx}} (\frac{1}{t} - \frac{t}{1 + t^{2}}) dt = {[\ln ∣ t ∣]}_{\frac{1}{e}}^{\frac{1}{tanx}} - {[\frac{1}{2} \ln (1 + t^{2})]}_{\frac{1}{e}}^{\frac{1}{tanx}}$ $= - \ln ∣ tanx ∣ + 1 - \frac{1}{2} \ln (1 + \frac{1}{\tan^{2} x}) - \frac{1}{2} \ln (1 + e^{- 2})$ $= - \ln ∣ tanx ∣ - \frac{1}{2} \ln (1 + \tan^{2} x) + \ln ∣ tanx ∣ - \frac{1}{2} \ln (1 + e^{- 2}) + 1 \Rightarrow$ $\int_{\frac{1}{e}}^{tanx} \frac{tdt}{1 + t^{2}} + \int_{\frac{1}{e}}^{cotanx} \frac{dt}{t (1 + t^{2})} = \frac{1}{2} \ln (1 + \tan^{2} x) - \frac{1}{2} \ln (1 + e^{- 2})$ $- \frac{1}{2} \ln (1 + \tan^{2} x) - \frac{1}{2} \ln (1 + e^{- 2}) + 1 = 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum