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nexus9112 [7]
3 years ago
11

Integrate Sec (4x - 1) Tan (4x - 1) Dx ​

Mathematics
1 answer:
Delicious77 [7]3 years ago
5 0

\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

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3 years ago
What is the slope of y = 1 + 5?
IgorLugansk [536]

Step-by-step explanation:

if there is no typo, then that is

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2 years ago
If DF=78, DE=5x-9, and EF=2x+10, find DE.
Kazeer [188]

Given that E is a point between Point D and F, the numerical value of segment DE is 46.

<h3>What is the numerical value of DE?</h3>

Given the data in the question;

  • E is a point between point D and F.
  • Segment DF = 78
  • Segment DE = 5x - 9
  • Segment EF = 2x + 10
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Since E is a point between point D and F.

Segment DF = Segment DE + Segment EF

78 = 5x - 9 + 2x + 10

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Hence,

Segment DE = 5x - 9

Segment DE = 5(11) - 9

Segment DE = 55 - 9

Segment DE = 46

Given that E is a point between Point D and F, the numerical value of segment DE is 46.

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Answer:

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