Biology: 11.1 Ferns mostly occur inenvironment.11.2generation is the dominantIn Pteridophytes thegeneration.11.3 The adult fern plant represents thegeneration.11.4develop from the rhizome, anchor the plantin the soil and absorb water and mineral salts.and therefore produces spores over11.5 The sporophyte ismany generations.11.6 When a spore germinates it gives rise to a green, heart shapedthat represents thegeneration.to move to the ovum in the female sex11.7 Released sperm needorgan.11.8 After fertilization

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11.1 Ferns mostly occur inenvironment.11.2generation - id11111211758404, jadeevasquez200

Ответ:

ryliesheab

30.10.2020

Biology

$\large\textsf{Evaluate the integral}$

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

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

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

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

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

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

____________

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

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

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

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

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

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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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