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Question Number 122708 by mathdave last updated on 19/Nov/20

$$solve$$$${\int }_0^{\frac{\pi }{4}}\ln (2+{\tan }^{2}x)dx$$

Commented by Dwaipayan Shikari last updated on 19/Nov/20

$${\int }_0^{\frac{\pi }{4}}log(2+{tan}^{2}x)dxtanx=\sqrt{2}t\Rightarrow {sec}^{2}x=\sqrt{2}\frac{dt}{dx}$$$${\int }_0^{\frac{1}{\sqrt{2}}}\frac{log(2+2t^2)}{2t^2+1}dt$$$$=\sqrt{2}{\int }_0^{\frac{1}{\sqrt{2}}}\frac{log\left(2\right)}{2t^2+1}+\frac{log(1+t^2)}{2t^2+1}dt$$$$={\left[log\left(2\right){tan}^{-1}\sqrt{2}t\right]}_0^{\frac{1}{\sqrt{2}}}+\sqrt{2}{\int }_0^{\frac{1}{\sqrt{2}}}\frac{log(1+t^2)}{2t^2+1}dt$$$$=\frac{\pi }{4}log\left(2\right)+\sqrt{2}I\left(a\right)$$$$I\left(a\right)={\int }_0^{\frac{1}{\sqrt{2}}}\frac{log(1+{at}^2)}{2t^2+1}dt$$$$I^{\prime }\left(a\right)={\int }_0^{\frac{1}{\sqrt{2}}}\frac{t^2}{(2t^2+1)({at}^2+1)}dt$$$$I^{\prime }\left(a\right)=\frac{1}{a-2}{\int }_0^{\frac{1}{\sqrt{2}}}\frac{1}{2t^2+1}-\frac{1}{{at}^2+1}dt$$$$I^{\prime }\left(a\right)=\frac{1}{\sqrt{2}(a-2)}{\left[{tan}^{-1}\sqrt{2}t\right]}_0^{\frac{1}{\sqrt{2}}}-{\left[\frac{1}{\sqrt{a}(a-2)}{tan}^{-1}\sqrt{a}t\right]}_0^{\frac{1}{\sqrt{2}}}$$$$I^{\prime }\left(a\right)=\frac{\pi }{4\sqrt{2}(a-2)}-\frac{{tan}^{-1}\sqrt{\frac{a}{2}}}{\sqrt{a}(a-2)}$$$$I\left(a\right)=\frac{1}{\sqrt{2}}\int \frac{\pi }{4(a-2)}-\frac{{tan}^{-1}\sqrt{\frac{a}{2}}}{\sqrt{\frac{a}{2}}(a-2)}.....$$

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