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Imagine if I have to evaluate the integral $\int\frac{7}{16+9r^2}dr$.

I have determined $u=\frac34r$ and $du=\frac34dr$. Now I have found the right steps to be: $$\int\frac7{16+9r^2}dr=\frac7{16}\int\frac1{1+\frac{9}{16}r^2}dr\\=\frac7{12}\int\frac1{1+(\frac34r)^2}(\frac34dr)\\=\frac7{12}\int\frac1{1+u^2}du\\=\frac7{12}tan^{-1}\frac34r+C$$

My question is that I do not understand how it goes from $\frac7{16}$ on the first line to $\frac7{12}$ on the second line? Any easy to understand explanation will be helpful as I am completely lost.

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  • $\begingroup$ note the $3/4$ in the differential and use $7/16=7/12\times 3/4$ $\endgroup$ – David Jaramillo Jan 27 '18 at 23:12
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    $\begingroup$ $7/16=3/4 \times 7/12$ $\endgroup$ – Naj Kamp Jan 27 '18 at 23:12
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Just as another way to see it, you could rewrite the relationship between $u$ and $r$ as $r = \frac{4}{3} u$ and $dr = \frac{4}{3} du$. Then you would have $$ \frac7{16}\int\frac1{1+\frac{9}{16}r^2}dr = \frac7{16}\int\frac1{1+\frac{9}{16}r^2}\left(\frac{4}{3} du \right) = \frac7{16} \frac{4}{3} \int\frac1{1+u^2}du = \frac7{12} \int\frac1{1+u^2}du. $$

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  • $\begingroup$ Also how is the u part of tan^3xsec^2x tanx? $\endgroup$ – Rixwaan Jan 27 '18 at 23:49
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The reason is that $$ \frac{7}{16} = \frac{7}{12}\frac{3}{4}.$$ They have brought the $\frac{3}{4}$ inside the integral next to the $dr$ in preparation for the substitution $u = \frac{3}{4}r.$

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$$I=\int\frac7{16+9r^2}dr=\frac7{16}\int\frac1{1+\frac{9}{16}r^2}dr$$

Substitute $u= \frac 34 r \text{ and } du=\frac 34 dr$

$$I=\frac7{16}\frac 43\int\frac {du}{1+u^2}=\frac7{12}\int\frac {du}{1+u^2}$$

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