In the period immediately prior to World War I, Europe control percent of the world's territory.

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ryliesheab

30.10.2020

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$\large\textsf{Evaluate the integral}$

$\large\begin{array}{l} \mathsf{\displaystyle\int\!tan^{-1}(4x)\,dx\qquad\quad(i)} \end{array}$

$\large\begin{array}{l} \textsf{I think a way to start this evaluation is substituting}\\\\ \mathsf{tan^{-1}(4x)=t\qquad\left(-\,\dfrac{\pi}{2}<t<\dfrac{\pi}{2}\right)}\\\\\\ \textsf{then,}\\\\ \mathsf{4x=\mathsf{tan\,}t}\\\\ \mathsf{x=\dfrac{1}{4}\,\mathsf{tan\,}t~~\Rightarrow~~dx=\dfrac{1}{4}\,sec^2\,t\,dt} \end{array}$

$\large\begin{array}{l} \textsf{so that (i) becomes}\\\\ \mathsf{\displaystyle\int\!t\cdot \frac{1}{4}\,sec^2\,t\,dt}\\\\ =\mathsf{\displaystyle\int\! \frac{1}{4}\,t\,sec^2\,t\,dt\qquad\quad(ii)} \end{array}$

$\large\begin{array}{l} \textsf{Now integrate (ii) by parts:}\\\\ \begin{array}{lcl} \mathsf{u=\dfrac{1}{4}\,t}&~\Rightarrow~&\mathsf{du=\dfrac{1}{4}\,dt}\\\\ \mathsf{dv=sec^2\,t}&~\Leftarrow~&\mathsf{v=tan\,t} \end{array} \end{array}$

$\large\begin{array}{l} \textsf{So,}\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\ \mathsf{\displaystyle\int\!\dfrac{1}{4}\,t\,sec^2\,t\,dt=\frac{1}{4}\,t\cdot tan\,t-\int\!tan\,t\cdot \frac{1}{4}\,dt}\\\\ \mathsf{\displaystyle\int\!\dfrac{1}{4}\,t\,sec^2\,t\,dt=\frac{1}{4}\,t\cdot tan\,t-\frac{1}{4}\int\!\frac{sin\,t}{cos\,t}\,dt}\\\\ \mathsf{\displaystyle\int\!\dfrac{1}{4}\,t\,sec^2\,t\,dt=\frac{1}{4}\,t\cdot tan\,t+\frac{1}{4}\int\!\frac{1}{cos\,t}\cdot (-\,sin\,t)\,dt} \end{array}$

$\large\begin{array}{l} \mathsf{\displaystyle\int\!\dfrac{1}{4}\,t\,sec^2\,t\,dt=\frac{1}{4}\,t\cdot tan\,t+\frac{1}{4}\int\!\frac{1}{w}\,dw\qquad(w=cos\,t)}\\\\ \mathsf{\displaystyle\int\!\dfrac{1}{4}\,t\,sec^2\,t\,dt=\frac{1}{4}\,t\cdot tan\,t+\frac{1}{4}\,\ell n|w|+C}\\\\ \mathsf{\displaystyle\int\!\dfrac{1}{4}\,t\,sec^2\,t\,dt=\frac{1}{4}\,t\cdot tan\,t+\frac{1}{4}\,\ell n|cos\,t|+C\qquad\quad(iii)} \end{array}$

____________

$\large\begin{array}{l} \textsf{Now how to get back to the x variable?}\\\\ \textsf{It becomes simple, just by using trig identities. :-)}\\\\\\ \textsf{Remember that}\\\\ \mathsf{x=\dfrac{1}{4}\,tan\,t\qquad\quad\left(-\,\dfrac{\pi}{2}<t<\dfrac{\pi}{2}\right)}\\\\ \mathsf{x=\dfrac{1}{4}\cdot \dfrac{sin\,t}{cos\,t}}\\\\ \mathsf{x^2=\left(\dfrac{1}{4}\cdot \dfrac{sin\,t}{cos\,t}\right)^{\!2}} \end{array}$

$\large\begin{array}{l} \mathsf{x^2=\dfrac{1}{16}\cdot \dfrac{sin^2\,t}{cos^2\,t}}\qquad\quad\textsf{(but }\mathsf{sin^2\,t=1-cos^2\,t}\textsf{)}\\\\ \mathsf{x^2=\dfrac{1}{16}\cdot \dfrac{1-cos^2\,t}{cos^2\,t}}\\\\ \mathsf{16x^2=\dfrac{1-cos^2\,t}{cos^2\,t}}\\\\ \mathsf{16x^2=\dfrac{1}{cos^2\,t}-\dfrac{cos^2\,t}{cos^2\,t}} \end{array}$

$\large\begin{array}{l} \mathsf{16x^2=\dfrac{1}{cos^2\,t}-1}\\\\ \mathsf{16x^2+1=\dfrac{1}{cos^2\,t}}\\\\ \mathsf{\dfrac{1}{cos\,t}=\sqrt{16x^2+1}}\\\\ \mathsf{cos\,t=\dfrac{1}{\sqrt{16x^2+1}}} \end{array}$

$\large\begin{array}{l} \textsf{Then (iii) becomes}\\\\ \mathsf{\displaystyle\int\!tan^{-1}(4x)\,dx=\frac{1}{4}\,tan^{-1}(4x)\cdot 4x+\frac{1}{4}\,\ell n\left|\dfrac{1}{\sqrt{16x^2+1}}\right|+C}\\\\\\ \textsf{Simplifying...}\\\\ \mathsf{\displaystyle\int\!tan^{-1}(4x)\,dx=\frac{1}{\diagup\!\!\!\! 4}\,tan^{-1}(4x)\cdot \diagup\!\!\!\! 4x+\frac{1}{4}\,\ell n\Big[\left(16x^2+1\right)^{\!\!-\frac{1}{2}}\Big]+C}\\\\ \mathsf{\displaystyle\int\!tan^{-1}(4x)\,dx=x\cdot tan^{-1}(4x)+\frac{1}{4}\cdot \left(-\dfrac{1}{2}\right)\,\ell n\!\left(16x^2+1\right)+C} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int\!tan^{-1}(4x)\,dx=x\cdot tan^{-1}(4x)-\frac{1}{8}\,\ell n(16x^2+1)+C} \end{array}}\qquad\checkmark \end{array}$

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$\large\textsf{I hope this helps. :-)}$

Tags: indefinite integral integrate inverse tangent arctan trigonometric substituition parts trig identity integral calculus

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