Evaluate the following trigonometric integral. \displaystyle\int 8\operatorname { cos } ^ { 3 } 4...

Question:

Evaluate the following trigonometric integral.

{eq}\displaystyle\int 8\operatorname { cos } ^ { 3 } 4 x d x {/eq}

Indefinite Integration by U-Substitution:

In this integration, we will apply the technique of u-substitution but before applying this technique, we will simplify the integrand into a more simpler and integrable form. The following trigonometric identity will be applied to simplify the integrand:

{eq}\sin^2\theta+\cos^2\theta=1 {/eq}

Answer and Explanation:

{eq}\begin{align} \int 8 \cos^3 (4 x) d x &= 8\int \cos^2 (4 x) \cos (4 x) d x\\ &= 8\int \left[ 1-\sin^2 (4 x) \right] (4\cos (4 x) d x) \\ &= 2\int \left[ 1-u^2\right] \, du & \left[\text{ Apply the following u-substitution:} \matrix{u &=\sin (4 x)\\ \Rightarrow du& = 4\cos (4 x) d x } \right]\\ &= 2 \left[ u-\frac{u^3}{3} \right]+C & \left[\text{ Integrate with respect to} \, u\right]\\ &= 2\sin (4 x) -\frac{2\sin^3 (4 x)}{3} +C & \left[\text{ Reverse the substitution } \right]\\ \end{align} {/eq}

Therefore, {eq}\displaystyle \boxed{\color{blue} { \int 8 \cos^3 (4 x) d x = 2\sin (4 x) -\frac{2\sin^3 (4 x)}{3} +C }} {/eq}


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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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