Question

Integrate with respect to t: ∫ sec^2 x dx

a. sinx
b. sinx + c
c. sec x + c
d. tanx + c

Answer 1

Integrate with respect to $t$: $\int \sec^2 x \, dx = \int \frac{1}{\cos^2(x)} dx = \int \frac{1}{\cos^2(x)} dx = \int \frac{(\sin x)^2}{\cos^2(x)} dx = \int \frac{(\sin x)^2}{\cos(x)} dx = \int \frac{(\cos x)^2}{\cos(x)} dx = \int \frac{(\sin x)^2}{\cos(x)} dx = \frac{\sin x}{2x} dx.$

we can use that $1 = \sin^2(x) + \cos^2(x)$ and get:

$$= \int \frac{\sin^2(x) + \cos^2(x)}{\cos^2(x)} dx = \int \frac{-\left(-\sin(x)\right) * \sin x + \cos(x) * \cos(x)}{\cos^2(x)} dx$$

but $(\cos(x))' = -\sin(x)$, $(\sin(x))' = \cos(x)$ and we get:

$$= \int \frac{-\left(\cos(x)\right)' * \sin(x) + \left(\sin(x)\right)' * \cos(x)}{\cos^2(x)} dx = \int \left(\frac{\sin x}{\cos x}\right)' dx = \frac{\sin x}{\cos x} + c = \tan x + c$$

here we use rule of fraction differentiation $\frac{-\left(\cos(x)\right)' * \sin(x) + \left(\sin(x)\right)' * \cos(x)}{\cos^2(x)} = \left(\frac{\sin x}{\cos x}\right)'$

So correct answer D.