MA005: Calculus I

Practice 1: $\begin{aligned} &\mathbf{D}\left(\tan ^{2}(x)+\mathrm{C}\right)=2 \tan ^{1}(x) \mathbf{D}(\tan (x))=2 \tan (x) \sec ^{2}(x) \\ &\mathbf{D}\left(\sec ^{2}(x)+\mathrm{C}\right)=2 \sec ^{1}(x) \mathbf{D}(\sec (x))=2 \sec (x) \sec (x) \tan (x)=2 \tan (x) \sec ^{2}(x) \end{aligned}$

Practice 2: We know $\mathbf{D}(\tan (x))=\sec ^{2}(x)$, so it is reasonable to try $\tan (7 x)$.

$\mathbf{D}(\tan (7 x))=\sec ^{2}(7 x) \mathbf{D}(7 x)=7 \cdot \sec ^{2}(7 x)$, a result 7 times the function we want, so

divide the original "guess" by 7 and try it. $\mathrm{D}\left(\frac{1}{7} \tan (7 x)\right)=\frac{1}{7} \cdot 7 \cdot \sec ^{2}(7 x)=\sec ^{2}(7 x)$.

$\int \sec ^{2}(7 x) \mathrm{dx}=\frac{1}{7} \tan (7 x)+\mathrm{C}$

$\mathbf{D}\left((3 x+8)^{1 / 2}\right)=\frac{1}{2} \cdot(3 x+8)^{-1 / 2} \mathbf{D}(3 x+8)=\frac{3}{2} \cdot(3 x+8)^{-1 / 2}$ so let's multiply our original "guess" by 2/3:

$\mathbf{D}\left(\frac{2}{3}(3 x+8)^{1 / 2}\right)=\frac{2}{3} \frac{1}{2} \cdot(3 x+8)^{-1 / 2} \mathbf{D}(3 x+8)=\frac{2}{3} \frac{3}{2} \cdot(3 x+8)^{-1 / 2}=(3 x+8)^{-1 / 2}$

$\int \frac{1}{\sqrt{3 x+8}} \mathrm{dx}=\frac{2}{3}(3 x+8)^{1 / 2}+\mathrm{C}$.

Practice 3:

(a) $\int(7 x+5)^{3} \mathrm{dx}$ with $u=7 x+5$. Then $\mathrm{du}=7 \mathrm{dx}$ so $\mathrm{dx}=\frac{1}{7} \mathrm{du}$ and $\int(7 x+5)^{3} \mathrm{dx}=\int \mathrm{u}^{3} \frac{1}{7} \mathrm{du}=\frac{1}{7} \cdot \frac{1}{4} \mathrm{u}^{4}+\mathrm{C}=\frac{1}{28}(7 x+5)^{4}+\mathrm{C}$.

(b)$\int \sin \left(x^{3}-1\right) 3 x^{2} \mathrm{dx}$ with $u=x^{3}-1$. Then $\mathrm{du}=3 x^{2} \mathrm{dx}$ so

$\int \sin \left(x^{3}-1\right) 3 x^{2} \mathrm{dx}=\int \sin (u) \mathrm{du}=-\cos (u)+\mathrm{C}=-\cos \left(x^{3}-1\right)+\mathrm{C}$.

Practice 4:

(a) $\mathrm{u}=2+\sin (3 \mathrm{x})$ so when $\mathrm{x}=0, \mathrm{u}=2+\sin (3+0)=2$. When $\mathrm{x}=\pi$, $\mathrm{u}=2+\sin (3 \cdot \pi)=2$. (This integral is now very easy to evaluate. Why?)

(b) $\mathrm{u}=2+\mathrm{e}^{\mathrm{x}}$ so when $\mathrm{x}=0, \mathrm{u}=2+\mathrm{e}^{0}=3$. When $\mathrm{x}=2, \mathrm{u}=2+\mathrm{e}^{2}$.

The new integration endpoints are $\mathrm{u}=3$ to $\mathrm{u}=2+\mathrm{e}^{2}$.

(c) $\mathrm{u}=\mathrm{e}^{\mathrm{x}}$ so when $\mathrm{x}=0, \mathrm{u}=\mathrm{e}^{0}=1$. When $\mathrm{x}=\ln (3), \mathrm{u}=\mathrm{e}^{\ln (3)}=3$.

The new integration endpoints are $\mathbf{u}=1$ to $\mathbf{u}=3$.