MA005: Calculus I

Antiderivatives are very sensitive to small changes in the integrand, and we need to be very careful. Fortunately, an antiderivative can always be checked by differentiating, so even though we may not find the correct antiderivative, we should be able to determine if an antiderivative is incorrect.

Example 1: We know one antiderivative of $\cos (x)$ is $\sin (x)$, so $\int \cos (x) \mathrm{dx}=\sin (x)+\mathrm{C}$. Find $(a) \int \cos (2 x+3) \mathrm{dx}, (b) \int \cos (5 x-7) \mathrm{dx}, and (c) \int \cos \left(x^{2}\right) \mathrm{dx}.$

Solution:

(a) Since $\sin (x)$ is an antiderivative of $\cos (x)$, it is reasonable to hope that $\sin (2 x+3)$ will be an antiderivative of $\cos (2 x+3)$. When we differentiate $\sin (2 x+3)$, however, we see that $\mathbf{D}(\sin (2 x+3))=\cos (2 x+3) \cdot 2$, exactly twice the result we want. Let's try again by modifying our "guess" and differentiating. Since the result of differentiating was twice the function we wanted, lets try one half the original function $\mathbf{D}\left(\frac{1}{2} \sin (2 x+3)\right)=\frac{1}{2} \cos (2 x+3) \cdot 2=\cos (2 x+3)$, exactly what we want. Therefore, $\int \cos (2 x+3) \mathrm{dx}=\frac{1}{2} \sin (2 x+3)+\mathrm{C}$.

(b) $\mathbf{D}(\sin (5 x-7))=\cos (5 x-7) \cdot \mathbf{5}$, but $\mathbf{D}\left(\frac{1}{5} \sin (5 x-7)\right)=\frac{1}{5} \cos (5 x-7) \cdot 5=\cos (5 x-7)$ so $\int \cos (5 x-7) \mathrm{dx}=\frac{1}{5} \sin (5 x-7)+\mathrm{C}$.

(c) $\mathbf{D}\left(\sin \left(x^{2}\right)\right)=\cos \left(x^{2}\right) \cdot 2 x$. It was easy enough in parts (a) and (b) to modify our "guesses" to eliminate the constants 2 and 5, but the $x$ is much harder to eliminate.

$\mathbf{D}\left(\frac{1}{2 x} \sin \left(x^{2}\right)\right)=\mathbf{D}\left(\frac{\sin \left(x^{2}\right)}{2 x}\right)$ requires the quotient rule (or the product rule), and

$\mathbf{D}\left(\frac{\sin \left(x^{2}\right)}{2 x}\right)=\frac{2 x \cdot \mathbf{D}\left(\sin \left(x^{2}\right)\right)-\sin \left(x^{2}\right) \cdot \mathbf{D}(2 x)}{(2 x)^{2}}=\cos \left(x^{2}\right)-\frac{\sin \left(x^{2}\right)}{2 x^{2}}$.

which is not the result we want. A fact which can be proved using more advanced mathematics is that $\cos \left(x^{2}\right)$ does not have an "elementary" antiderivative composed of polynomials, roots, trigonometric functions, or exponential functions or their inverses. The value of a definite integral of $\cos \left(x^{2}\right)$ could still be approximated as accurately as needed by using Riemann sums or one of the numerical techniques in Section 4.9 . The point of part (c) is that even a simple looking integrand can be very difficult. At this point, there is no easy way to tell.