$\bf \displaystyle\int~sec(4x-1)tan(4x-1)dx \\\\[-0.35em] ~\dotfill\\\\ sec(4x-1)tan(4x-1)\implies \cfrac{1}{cos(4x-1)}\cdot \cfrac{sin(4x-1)}{cos(4x-1)}\implies \cfrac{sin(4x-1)}{cos^2(4x-1)} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int~\cfrac{sin(4x-1)}{cos^2(4x-1)}dx \\\\[-0.35em] ~\dotfill\\\\ u=cos(4x-1)\implies \cfrac{du}{dx}=-sin(4x-1)\cdot 4\implies \cfrac{du}{-4sin(4x-1)}=dx \\\\[-0.35em] ~\dotfill$

$\bf \displaystyle\int~\cfrac{~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{u^2}\cdot \cfrac{du}{-4~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies -\cfrac{1}{4}\int\cfrac{1}{u^2}du\implies -\cfrac{1}{4}\int u^{-2}du \\\\\\ -\cfrac{1}{4}\cdot \cfrac{u^{-2+1}}{-1}\implies \cfrac{1}{4}\cdot u^{-1}\implies \cfrac{1}{4u}\implies \cfrac{1}{4cos(4x+1)}+C$

5 0

3 years ago

Rema spent 6 days reading her book, completing the same fraction of the book each day, if she read a total of 8/9 of her book af

AnnZ [28]

(8/9) / 6 =
8/9 * 1/6 =
8/54 which reduces to 4/27 of the book each day <===

7 0

3 years ago