### Calculus

Topics within Calculus

## Practice

Now that you've had some time to play around with antiderivatives, it's time to see what you can accomplish with them!

• $\displaystyle{\int \sec^{2} x \ dx = }$ $+C$

What function do we differentiate to get $\sec^{2} x$ ?

We know that $$\frac{d \tan x + C}{dx} = \sec^{2} x.$$ Therefore, $$\int sec^{2} x dx = \tan x +C.$$

• $\displaystyle{\int \sec x \tan x \ dx = }$ $+C$

What function do we differentiate to get $\sec x \tan x$ ?

We know that $$\frac{d \sec x + C}{dx} =\sec x \tan x.$$ Therefore, $$\int \sec x \tan x x dx = \sec x +C.$$

• $\displaystyle{ \int e^{x} dx= }$ $+C$

Try differentiating the function instead. What do you notice or remember about $e^x$?

We know that $$\frac{d e^x + C}{dx}= e^x .$$ So, we can say that $$\int e^x dx = e^x + C.$$

• $\displaystyle{\frac{1}{n} \int n\cdot x^{n-1} dx = }$ $+C$

Evaluate $$\frac{d x^n}{dx}.$$ How can you use this information?

We know that $$\frac{d x^n}{dx} = n\cdot x^{n-1}.$$ We can use this to conclude that $$\int n\cdot x^{n-1} dx = x^n+C.$$ But wait! We also had to worry about a fraction. So, we multiply both sides by $\frac{1}{n}$, and we get $$\frac{x^n}{n} +C.$$ (Note that " $+C$ " represents an arbitrary constant, and " $C/n$"  is still a constant. Hence, we just write "$+C.$")