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Found in: Page 326

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the Integral: $$\int {{{\cos }^2}} x\sin 2xdx$$

To evaluate the integral $$\int {{{\cos }^2}} x\sin 2xdx$$, we will use the identity $${\cos ^2}x = \frac{{1 + \cos 2x}}{2}$$ followed by integration formulas

$$\int {{{\cos }^2}} x\sin 2xdx = \frac{{ - 1}}{4}\cos 2x - \frac{1}{{16}}\cos 4x + c$$

See the step by step solution

## Step 1: Using trigonometric double angle formula

Let I = $$\int {{{\cos }^2}} x\sin 2xdx$$

As $${\cos ^2}x = \frac{{1 + \cos 2x}}{2}$$

$$I = \int {(\frac{{1 + \cos 2x}}{2}} )\sin 2xdx$$

## Step 2: Multiplying functions and splitting of integrals

\begin{aligned}{l}I &= \frac{1}{2}\int {(1 + \cos 2x)\sin 2xdx} \\ &= \frac{1}{2}\int {(\sin 2x + \sin 2x\cos 2x)dx} \end{aligned}

Now split the integral:

$$I = \frac{1}{2}\int {\sin 2xdx + \frac{1}{2}} \int {\sin 2x\cos 2xdx}$$

(1 Term) (2 Term)

## Step 3: Multiply and divide 2 terms by 2 and use identity $$2\sin 2x\cos 2x - \sin 4x$$

\begin{aligned}{l}I &= \frac{1}{2}\int {\sin 2xdx + \frac{1}{4}} \int {2\sin 2x\cos 2xdx} \\ &= \frac{1}{2}\int {\sin 2xdx + \frac{1}{4}} \int {\sin 4xdx} \\ &= \frac{1}{2}(\frac{{ - 2\cos x}}{2}) + \frac{1}{4}(\frac{{ - \cos 4x}}{4}) + c\\\end{aligned}

$$I = \frac{{ - \cos 2x}}{4} - \frac{1}{{16}}\cos 4x + c$$

Hence value of integral:

$$\int {{{\cos }^2}} x\sin 2xdx = \frac{{ - 1}}{4}\cos 2x - \frac{1}{{16}}\cos 4x + c$$