Mathematics, 02.11.2019 03:31 angel234wilcox

Find the antiderivative for each function when c = 0. check your answers by differentiation.

(a) h(x)= (secant x)^2
(b) g(x)= 8/9(secant x/9)^2
(c) k(x)= (- secant 9x/8)^2

Answers

Answer from: IHeartDarkSide03

Answer with Step-by-step explanation:

We are given that C=0

We have to find the anti-derivative of each function

a.h(x)=sec^2 x

We know that $\int sec^2 xdx=tanx+C$

Apply this formula then, we get

$\int sec^2x dx=tanx +C$

Substitute C=0

Then, we get $\int h(x) dx=tan x$

Verification :

Differentiate w.r.t x

Then, we get

h(x)=sec^2 x

$\frac{d tanx}{dx}=sec^2 x$

$b.g(x)=\frac{8}{9} sec^2 \frac{x}{9}$

$\int g(x) dx=\frac{8}{9}\int sec^2\frac{x}{9} dx=\frac{8}{9}\times 9tan\frac{x}{9}+C$

Substitute C=0

$\int g(x) dx=8 tan\frac{x}{9}$

Verification:

Differentiate w.r.t x

$g(x)=8\times \frac{1}{9} sec^2\frac{x}{9}=\frac{8}{9} sec^2\frac{x}{9}$

$c.k(x)=sec^2\frac{9x}{8}$

$\int k(x) dx=\int \sec^28}{9}tan\frac{9x}{8}+C$

Substitute C=0

$\int k(x) dx=\frac{8}{9} tan\frac{9x}{8}$

Verification: Differentiate w.r.t x

$k(x)=\frac{8}{9}\times \frac{9}{8} sec^2\frac{9x}{8}=sec^2\frac{9x}{8}$

Answer from: Quest

astronomers is your final answer! : )

Which professionals most directly use geometry in their work?

Answer from: Quest

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post
Can someone plz me 1 to 37 . i will earn 10 points

Can someone plz me 1 to 37 . i will earn 10 points

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Find the antiderivative for each function when c = 0. check your answers by differentiation.
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