Question

#39 how to do the integral

Answer 1

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Finding y by integration:

    
37.  $\mathsf{\dfrac{dy}{dx}=sin\,3x}$


Integrate both sides, and you get

$\mathsf{\displaystyle\int\!\dfrac{dy}{dx}\,dx=\int\! sin\,3x\,dx}\\\\\\ \mathsf{\displaystyle y=\int\! \frac{1}{3}\cdot 3\,sin\,3x\,dx}\\\\\\ \mathsf{\displaystyle y=\frac{1}{3}\int\! sin\,3x\cdot 3\,dx\qquad\quad(i)}$


Now, make a substitution:

$\mathsf{3x=u\quad\Rightarrow\quad 3\,dx=du}$


so the integral (i) becomes

$\mathsf{\displaystyle y=\frac{1}{3}\int\! sin\,u\,du}\\\\\\ \mathsf{y=\dfrac{1}{3}\cdot (-\,cos\,u)+C}\\\\\\ \mathsf{y=-\,\dfrac{1}{3}\,cos\,3x+C\qquad\quad\checkmark}$

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39.  $\mathsf{\dfrac{dy}{dx}=x\sqrt{x-3}}$

$\mathsf{\displaystyle\int\!\dfrac{dy}{dx}\,dx=\int\! x\sqrt{x-3}\,dx}}\\\\\\ \mathsf{\displaystyle y=\int\! x\sqrt{x-3}\,dx\qquad\quad(ii)}$


Substitution:

$\begin{array}{lcl} \mathsf{\sqrt{x-3}=u}&\quad\Rightarrow\quad&\mathsf{x-3=u^2}\\\\ &&\mathsf{x=u^2+3\quad\Rightarrow\quad dx=2u\,du} \end{array}$


and the integral (ii) becomes

$\mathsf{\displaystyle y=\int\! (u^2+3)\cdot u\cdot 2u\,du}\\\\ \mathsf{\displaystyle y=\int\! (u^2+3)\cdot 2u^2\,du}$


Multiply with the distrubutive property, and then integrate it by using the power rule:

$\mathsf{\displaystyle y=\int\! (2u^4+6u^2)\,du}\\\\\\ \mathsf{y=2\cdot \dfrac{u^{4+1}}{4+1}+6\cdot \dfrac{u^{2+1}}{2+1}+C}\\\\\\ \mathsf{y=2\cdot \dfrac{u^5}{5}+6\cdot \dfrac{u^3}{3}+C}\\\\\\ \mathsf{y=\dfrac{2}{5}\,u^5+2u^3+C}$


Last, substitute back for u, so the answer is given in terms of x:

$\mathsf{y=\dfrac{2}{5}\cdot (\sqrt{x-3})^5+2\cdot (\sqrt{x-3})^3+C}\\\\\\ \mathsf{y=\dfrac{2}{5}\cdot \big[(x-3)^{1/2}\big]^5+2\cdot \big[(x-3)^{1/2}\big]^3+C}\\\\\\ \mathsf{y=\dfrac{2}{5}\,(x-3)^{5/2}+2(x-3)^{3/2}+C\qquad\quad\checkmark}$

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41.  $\mathsf{\dfrac{dy}{dx}=xe^{x^2}}$

$\mathsf{\displaystyle\int\! \frac{dy}{dx}\,dx=\int\!xe^{x^2}\,dx}\\\\\\ \mathsf{\displaystyle y=\int\!xe^{x^2}\,dx}\\\\\\ \mathsf{\displaystyle y=\int\!\frac{1}{2}\cdot 2xe^{x^2}\,dx}\\\\\\ \mathsf{\displaystyle y=\frac{1}{2}\int\! e^{x^2}\cdot 2x\,dx\qquad\quad(iii)}$


Substitution:

$\mathsf{x^2=u\quad\Rightarrow\quad 2x\,dx=du}$


so the integral (iii) becomes

$\mathsf{\displaystyle y=\frac{1}{2}\int\! e^u\,du}\\\\\\ \mathsf{\displaystyle y=\frac{1}{2}\cdot e^u+C}\\\\\\ \mathsf{\displaystyle y=\frac{1}{2}\,e^{x^2}+C\qquad\quad\checkmark}$


I hope this helps. =)


Tags:  indefinite integral u substitution differential integral calculus