Question Number 122708 by mathdave last updated on 19/Nov/20
$${solve} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{2}+\mathrm{tan}^{\mathrm{2}} {x}\right){dx} \\ $$
Commented by Dwaipayan Shikari last updated on 19/Nov/20
![∫_0 ^(π/4) log(2+tan^2 x)dx tanx=(√2)t⇒sec^2 x=(√2)(dt/dx) ∫_0 ^(1/( (√2))) ((log(2+2t^2 ))/(2t^2 +1))dt =(√2)∫_0 ^(1/( (√2))) ((log(2))/(2t^2 +1))+((log(1+t^2 ))/(2t^2 +1))dt =[log(2)tan^(−1) (√2)t]_0 ^(1/( (√2))) +(√2)∫_0 ^(1/( (√2))) ((log(1+t^2 ))/(2t^2 +1))dt =(π/4)log(2)+(√2)I(a) I(a)=∫_0 ^(1/( (√2))) ((log(1+at^2 ))/(2t^2 +1))dt I′(a)=∫_0 ^(1/( (√2))) (t^2 /((2t^2 +1)(at^2 +1)))dt I′(a)=(1/(a−2))∫_0 ^(1/( (√2))) (1/(2t^2 +1))−(1/(at^2 +1))dt I′(a)=(1/( (√2)(a−2)))[tan^(−1) (√2)t]_0 ^(1/( (√2))) −[(1/( (√a)(a−2)))tan^(−1) (√a)t]_0 ^(1/( (√2))) I′(a)=(π/( 4(√2)(a−2)))−((tan^(−1) (√(a/2)))/( (√a)(a−2))) I(a)=(1/( (√2)))∫(π/(4(a−2)))−((tan^(−1) (√(a/2)))/( (√(a/2))(a−2))).....](https://www.tinkutara.com/question/Q122746.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left(\mathrm{2}+{tan}^{\mathrm{2}} {x}\right){dx}\:\:\:\:\:\:\:\:\:{tanx}=\sqrt{\mathrm{2}}{t}\Rightarrow{sec}^{\mathrm{2}} {x}=\sqrt{\mathrm{2}}\frac{{dt}}{{dx}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \frac{{log}\left(\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \frac{{log}\left(\mathrm{2}\right)}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{log}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\left[{log}\left(\mathrm{2}\right){tan}^{−\mathrm{1}} \sqrt{\mathrm{2}}{t}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} +\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \frac{{log}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\frac{\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)+\sqrt{\mathrm{2}}{I}\left({a}\right) \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \frac{{log}\left(\mathrm{1}+{at}^{\mathrm{2}} \right)}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$${I}'\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \frac{{t}^{\mathrm{2}} }{\left(\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)\left({at}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\ $$$${I}'\left({a}\right)=\frac{\mathrm{1}}{{a}−\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{at}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$${I}'\left({a}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\left({a}−\mathrm{2}\right)}\left[{tan}^{−\mathrm{1}} \sqrt{\mathrm{2}}{t}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} −\left[\frac{\mathrm{1}}{\:\sqrt{{a}}\left({a}−\mathrm{2}\right)}{tan}^{−\mathrm{1}} \sqrt{{a}}{t}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \\ $$$${I}'\left({a}\right)=\frac{\pi}{\:\mathrm{4}\sqrt{\mathrm{2}}\left({a}−\mathrm{2}\right)}−\frac{{tan}^{−\mathrm{1}} \sqrt{\frac{{a}}{\mathrm{2}}}}{\:\sqrt{{a}}\left({a}−\mathrm{2}\right)} \\ $$$${I}\left({a}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\pi}{\mathrm{4}\left({a}−\mathrm{2}\right)}−\frac{{tan}^{−\mathrm{1}} \sqrt{\frac{{a}}{\mathrm{2}}}}{\:\sqrt{\frac{{a}}{\mathrm{2}}}\left({a}−\mathrm{2}\right)}….. \\ $$