A simple integral?

Using the substitution $u = \cos(x),$ we have that $du = -\sin(x) dx.$ The integral then becomes

$\displaystyle\int -\dfrac{1}{u^{3}} du = \dfrac{1}{2u^{2}} + c = \dfrac{\sec^{2}(x)}{2} + c.$

This can also be written as $\dfrac{\tan^{2}(x) + 1}{2} + c = \dfrac{\tan^{2}(x)}{2} + c,$ where the "extra" $\frac{1}{2}$ term was absorbed by the general constant $c.$ Thus both the explicit options provided are correct, since they differ only by a constant.

$∫ \frac{sin(x)}{cos^{3}(x)}dx$ $∫(\frac{sin(x)}{cos(x)})(\frac{1}{cos^{2}(x)})dx$ $∫tan(x)sec^{2}(x)dx$ Now, use U-substitution, but notice that two substitutions can be made $u=tan(x)$ $du = sec^{2}(x)dx$ Or $u=sec(x)$ $du = sec(x)tan(x)dx$ Either substitution will yield: $∫udu$ $∫udu = \frac{1}{2}u^{2} + c$ Depending on what substitution you made, the answer will either be: $\frac{1}{2}tan^{2}(x) + c$ Or $\frac{1}{2}sec^{2}(x) + c$

$\displaystyle \int \dfrac{\sin x}{\cos^3x} dx = \int \tan x \sec^2x dx$

Take $t = \tan x \Rightarrow dt = \sec^2x dx$

$\displaystyle \int t dt = \dfrac{t^2}{2} + c = \dfrac{\tan^2x}{2} + c = \dfrac{\sec^2x}{2} + c - \dfrac{1}{2} = \dfrac{\sec^2x}{2} + C$

The actual difference here is the constant of integration, but hence we do not know what it is we can simply assume it to be some $c$

use substitution : $\cos x = u \leftrightarrow -\sin x dx = du$

do the integral and you will get: $\frac{\sec ^2(x)}{2} + c$

wait a second... remember trigonometric identity: $\sec ^2 (x) = \tan ^2(x) + 1$

thus 2nd choice is also right.

that's why it's very important to reflect on your answer before submitting.

differentiate the options