9.) $int\tan^{2}bymathrm{d}y$

Explanation:

Step1: Use trigonometric identity

Recall that $\tan^{2}x=\sec^{2}x - 1$. So, $\int\tan^{2}(by)dy=\int(\sec^{2}(by)-1)dy$.

Step2: Split the integral

By the integral - sum rule $\int(f(x)+g(x))dx=\int f(x)dx+\int g(x)dx$, we have $\int(\sec^{2}(by)-1)dy=\int\sec^{2}(by)dy-\int 1dy$.

Step3: Integrate each part

For $\int\sec^{2}(by)dy$, let $u = by$, then $du = bdy$ and $\int\sec^{2}(by)dy=\frac{1}{b}\int\sec^{2}(u)du=\frac{1}{b}\tan(u)+C_1=\frac{1}{b}\tan(by)+C_1$. And $\int 1dy=y + C_2$.

Step4: Combine the results

$\int\tan^{2}(by)dy=\frac{1}{b}\tan(by)-y + C$, where $C = C_1 - C_2$.

Answer:

$\frac{1}{b}\tan(by)-y + C$