# reverse quotient rule and then integrating!!!!!! just not clicking!!!

• Apr 11th 2013, 03:55 PM
triffid1000
reverse quotient rule and then integrating!!!!!! just not clicking!!!
Hi, i have this initial-value problem....i need to use the reverse quotient rule, but i also need to use an integration formula too:

dy/dx= (cos(5x))/(7+sin(5x)) when y=6 when x=0

f'(x)/f(x)dx=in(f(x))+c (f(x)>0)

i don't want the answer i would just like to know how??? if that makes sense lol. I cannot find an example that is similar to this so i can work through it as i find this is the best way of learning....can anyone help pretty please??(Thinking)
• Apr 11th 2013, 06:39 PM
Soroban
Re: reverse quotient rule and then integrating!!!!!! just not clicking!!!
Hello, triffid1000!

Quote:

$\displaystyle \displaystyle f(x) \;=\;\int\frac{\cos5x\,dx}{7+\sin5x}\qquad f(0) = 6$

$\displaystyle \text{Let }\,u \,=\,7 + \sin5x \quad\Rightarrow\quad du \,=\,5\cos5x\,dx \quad\Rightarrow\quad \cos5x\,dx \,=\,\tfrac{1}{5}du$

$\displaystyle \displaystyle\text{Substitute:}\;f(x) \;=\;\int\frac{\frac{1}{5}\,du}{u} \;=\;\tfrac{1}{5}\int\frac{du}{u} \;=\;\tfrac{1}{5}\ln|u|+C$

$\displaystyle \text{Back-substitute: }\;f(x) \;=\;\tfrac{1}{5}\ln|7+\sin5x| + C$

$\displaystyle \text{Since }f(0) = 6,\,\text{ we have: }\:\tfrac{1}{5}\ln|7+\sin0| + C \:=\:6$

. . $\displaystyle \text{Hence: }\:C \:=\:6 - \tfrac{1}{5}\ln7$

$\displaystyle \text{Therefore: }\:f(x) \;=\;\tfrac{1}{5}\ln|7 + \sin5x| + 6 - \tfrac{1}{5}\ln7$

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This can be simplified beyond all recognition.

$\displaystyle \tfrac{1}{5}\ln|7+\sin5x| - \tfrac{1}{5}\ln7 + 6$

. . . $\displaystyle =\; \tfrac{1}{5}\bigg[\ln|7+\sin5x| - \ln7|\bigg] + 6$

. . . $\displaystyle =\; \tfrac{1}{5}\ln\left|\frac{7+\sin5x}{7}\right| + 6$

. . . $\displaystyle =\; \tfrac{1}{5}\ln\left|1 + \tfrac{1}{7}\sin5x\right| + 6$