# Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 29

$$-\frac{1}{12} \sin ^{3}(3 x) \cos (3 x)-\frac{1}{8} \sin (3 x) \cos (3 x)+\frac{3}{8} x+c$$

#### Work Step by Step

Given $$\int \sin ^{4}(3 x) d x$$ Let $$u=3x\ \ \Rightarrow \ du= 3dx$$ Then by using $$\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$$ \begin{align*} \int \sin ^{4}(3 x) d x&=\frac{1}{3} \int \sin ^{4} u d u\\ &=-\frac{1}{12} \sin ^{3} u \cos u+\frac{1}{4} \int \sin ^{2} u d u \quad(n=4)\\ &=-\frac{1}{12} \sin ^{3} u \cos u+\frac{1}{4}\left(-\frac{1}{2} \sin u \cos u+\frac{1}{2} \int d u\right) \quad(n=2)\\ &=-\frac{1}{12} \sin ^{3} u \cos u-\frac{1}{8} \sin u \cos u+\frac{1}{8} u+c\\ &=-\frac{1}{12} \sin ^{3}(3 x) \cos (3 x)-\frac{1}{8} \sin (3 x) \cos (3 x)+\frac{3}{8} x+c \end{align*}

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