• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q16E

Expert-verified
Essential Calculus: Early Transcendentals
Found in: Page 326
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

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 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:

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

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.