Indefinite integrals sanity check.

Question

I have a few indefinite integrals and antiderivatives that I just want to verify to myself. I basically want to show myself that the indefinite integral is indeed the antiderivative of the function that we are integrating.

Also is the best way to think about the indefinite integral just the antiderivative? Where did this notation come about? What was the motivation behind it?

1. Check to see if this equation is correct:

$$\int cos^2x \cdot dx = \frac{1}{2} x + \frac{1}{4} \sin{2x} + C$$

So my check:

$$\frac{d}{dx} \frac{1}{2}x + \frac{1}{4} \sin{2x} + C$$ $$= \frac{1}{2} + \frac{1}{4} \cos{2x} \cdot 2$$ $$= \frac{1}{2} + \frac{1}{2} \cos{2x}$$

I'm stuck here.

2. $$\int (x^2 + 1 + \frac{1}{x^2 + 1}) \, dx$$

$$= \int (x^2) + \int (1) + \int \frac{1}{x^2 + 1}$$

$$= \frac{x^3}{3} + x + \tan^1x + C$$

3. Sanity check:

$$\int \sec{t}(\sec{t} + \tan{t}))$$

$$=\int \sec^2{t} + \sec{t} \cdot \tan{t}$$

$$= \int \sec^2{t}\,dt + \int \sec{t} \cdot \tan{t}$$

$$ \tan{t} + \sec{t} + C$$

Answer 1

For 1) you may use the identity $$\cos(2x)=2\cos^2(x)-1$$

2) looks fine (I assume that is a typo and it should say $\tan^{-1}(x)$. Doing a sanity check will just lead directly back to the same expression as in integral.

3) is also fine. As with 2), differentiating will take you straight back to the integrand.

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Also is the best way to think about the indefinite integral just the antiderivative?

Yes, definitely. If you look at it like this, you can often do integrals by eye. For instance $$\int\sin x\cos^6 x\,dx=-\frac17\cos^7 x+C$$

Where did this notation come about? What was the motivation behind it?

The notation for the integral symbol? I believe that comes from the fact that an integral is just an area, so can be thought of as a sum of infinitely many strips. The integral symbol looks like a capital S, for "sum". This interpretation is related to the antiderivative of a function via the fundamental theorem of calculus.