Homework Help is Here – Start Your Trial Now!
Math
Calculus
Jo 17. cos² x sin? x dx

Jo 17. cos² x sin? x dx

Mathematics For Machine Technology
Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter58: Achievement Review—section Five
Section: Chapter Questions
Problem 30AR: Determine dimension x to 3 decimal places.
See similar textbooks
icon
Related questions
Question

Can slove 17 and 18 by this method?

 

EXAMPLE 3.4 An Integrand with an Even Power of Sine Evaluate f sinx dx. Solution Using the half-angle formula, we can rewrite the integral as sin x dx = 1- cos 2x) dx. We can evaluate this last integral by using the substitution u = 2x, so that du = 2 dx. This gives us sin? (1 – cos 2x)2 dx = (1 – cos u) du 1- cos u du - sin u) +c = (2x – sin 2x) + c. Since u 2.x. With some integrals, you may need to apply the half-angle formulas several times, as in example 3.5. EXAMPLE 3.5 An Integrand with an Even Power of Cosine Evaluate f cosx dx. Solution Using the half-angle formula for cosine, we have 小 1 (1+ cos 2x)? dx cos x dx = (cos x) dx = (1+ 2 cos 2x + cos 2x) dx. = Using the half-angle formula again, on the last term in the integrand, we get | cos“. dx = 1+2 cos 2x + =(1+ cos 4x) dx 3 1 sin 2x + sin 4x + c, 32
Transcribed Image Text:EXAMPLE 3.4 An Integrand with an Even Power of Sine Evaluate f sinx dx. Solution Using the half-angle formula, we can rewrite the integral as sin x dx = 1- cos 2x) dx. We can evaluate this last integral by using the substitution u = 2x, so that du = 2 dx. This gives us sin? (1 – cos 2x)2 dx = (1 – cos u) du 1- cos u du - sin u) +c = (2x – sin 2x) + c. Since u 2.x. With some integrals, you may need to apply the half-angle formulas several times, as in example 3.5. EXAMPLE 3.5 An Integrand with an Even Power of Cosine Evaluate f cosx dx. Solution Using the half-angle formula for cosine, we have 小 1 (1+ cos 2x)? dx cos x dx = (cos x) dx = (1+ 2 cos 2x + cos 2x) dx. = Using the half-angle formula again, on the last term in the integrand, we get | cos“. dx = 1+2 cos 2x + =(1+ cos 4x) dx 3 1 sin 2x + sin 4x + c, 32
Jo J-1/4 | cos x 18. (cos 17. sin? x dx x + sin x) dx 7/2 19. cos x sin' x dx 20. cot x csc* x dx -1/3 T/4 1 21. dx 22. dx r2,9 r216
Transcribed Image Text:Jo J-1/4 | cos x 18. (cos 17. sin? x dx x + sin x) dx 7/2 19. cos x sin' x dx 20. cot x csc* x dx -1/3 T/4 1 21. dx 22. dx r2,9 r216
Expert Solution
steps

Step by step

Solved in 3 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer