Question

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

Answer 1

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

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