\[\int \tan^{2}(3x) \, dx\]
1
Use Integration by Substitution.
Let u=3xu=3x, du=3 dxdu=3 \, dx, then dx=13 dudx=\frac{1}{3} \, du

2
Using uu and dudu above, rewrite ∫tan⁡2(3x) dx\int \tan^{2}(3x) \, dx.
∫tan⁡2u3 du\int \frac{\tan^{2}u}{3} \, du

3
Use Constant Factor Rule: ∫cf(x) dx=c∫f(x) dx\int cf(x) \, dx=c\int f(x) \, dx.
13∫tan⁡2u du\frac{1}{3}\int \tan^{2}u \, du

4
Use Pythagorean Identities: tan⁡2x=sec⁡2x−1\tan^{2}x=\sec^{2}x-1.
13∫sec⁡2u−1 du\frac{1}{3}\int \sec^{2}u-1 \, du

5
Use Sum Rule: ∫f(x)+g(x) dx=∫f(x) dx+∫g(x) dx\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx.
13(∫sec⁡2u du+∫−1 du)\frac{1}{3}(\int \sec^{2}u \, du+\int -1 \, du)

6
The derivative of tan⁡u\tan{u} is sec⁡2u\sec^{2}u.
13(tan⁡u+∫−1 du)\frac{1}{3}(\tan{u}+\int -1 \, du)

7
Use this rule: ∫a dx=ax+C\int a \, dx=ax+C.
tan⁡u−u3\frac{\tan{u}-u}{3}

8
Substitute u=3xu=3x back into the original integral.
tan⁡3x−3x3\frac{\tan{3x}-3x}{3}

9
Add constant.
tan⁡3x−3x3+C\frac{\tan{3x}-3x}{3}+C

Done
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