MA005: Calculus I

One method for finding antiderivatives is to "guess" the form of the answer, differentiate your "guess" and then modify your original "guess" so its derivative is exactly what you want it to be. Parts (a) and (b) of the previous example illustrate that technique.

Example 2: We know $\int \sec ^{2}(x) \mathrm{d} \mathrm{x}=\tan (x)+\mathrm{C}$ and $\int \frac{1}{\sqrt{x}} \mathrm{~d} \mathrm{x}=2 \sqrt{x}+\mathrm{C}$. Find (a) $\int \sec ^{2}(3 x+7) d x$ and (b) $\int \frac{1}{\sqrt{5 x+3}} \mathrm{dx}$.

Solution:

(a) If we "guess" an answer of $\tan (3 x+7)$ and then differentiate it, we get $\left.\mathbf{D}(\tan (3 x+7))=\sec ^{2}(3 x+7) \cdot \mathbf{D}(3 x+7)\right)=\sec ^{2}(3 x+7) \cdot 3$ which is 3 times too large. If we divide our original guess by 3 and try again, we have

$\mathbf{D}\left(\frac{1}{3} \tan (3 x+7)\right)=\frac{1}{3} \mathbf{D}(\tan (3 x+7))=\frac{1}{3} \sec ^{2}(3 x+7) \cdot 3=\sec ^{2}(3 x+7)$ so $\int \sec ^{2}(3 x+7) \mathrm{dx}=\frac{1}{3} \tan (3 x+7)+\mathrm{C}$.

(b) If we "guess" an answer of $2 \sqrt{5 x+3}$ and then differentiate it, we get $\mathbf{D}(2 \sqrt{5 x+3})=2 \cdot \frac{1}{2}(5 x+3)^{-1 / 2} \mathbf{D}(5 x+3)=5 \cdot(5 x+3)^{-1 / 2}=\frac{5}{\sqrt{5 x+3}}$ which is 5 times too large. Dividing our guess by 5 and differentiating, we have

$\mathbf{D}\left(\frac{2}{5} \sqrt{5 x+3}\right)=\frac{1}{5} \cdot \mathbf{D}(2 \sqrt{5 x+3})=\frac{1}{5} \cdot \mathbf{5} \cdot(5 x+3)^{-1 / 2}=\frac{1}{\sqrt{5 x+3}}$ $\int \frac{1}{\sqrt{5 x+3}} \mathrm{~d} x=\frac{2}{5} \sqrt{5 x+3}+\mathrm{C}$.

Practice 2: Find $\int \sec ^{2}(7 x) \mathrm{dx} \text { and } \int \frac{1}{\sqrt{3 x+8}} \mathrm{dx}$.

The "guess and check" method is a very effective technique if you can make a good first guess, one that misses the desired result only by a constant multiple. In that situation just divide the first guess by the unwanted constant multiple. If the derivative of your guess misses by more than a constant multiple, then more drastic changes are needed. Sometimes the next technique can help.