Register to reply 
∫〖Sin^(2 ) x cos2x dx〗 ( solving it, getting + c) could need some help 
Share this thread: 
#1
Jan2511, 12:52 AM

P: 1

1. The problem statement, all variables and given/known data
∫〖Sin^(2 ) x * cos2x dx〗 2. Relevant equations The right answer should be: sin 2x / 4  x/4  sin4x / 16 3. The attempt at a solution If i would set Sin^(2 ) x= (1cos2x)/2 I can replace it in the integral Then we get: ∫▒〖(1cos2x)/2 cos2x dx〗 As far as I am concerned we cannot use the constant multiple rule because there is no constant to multiply I don’t know how to go on from here to solve it 


#2
Jan2511, 01:14 AM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,683

OK, so you have
[tex]\begin{align*} \int \sin^2 x\cos 2x\,dx &= \int\left(\frac{1\cos 2x}{2}\right)\cos 2x\,dx \\ &= \frac{1}{2}\int(\cos 2x  \cos^2 2x)\,dx \\ &= \frac{1}{2}\int\cos 2x\,dx  \frac{1}{2}\int \cos^2 2x\,dx \end{align*}[/tex] Can you take it from there? Hint: use a trig identity for cos^{2} 2x. 


Register to reply 
Related Discussions  
∫∫ x^2 dA ; bounded by ellipse  Calculus & Beyond Homework  6  
∫sin(1/x) over [0,1]  Calculus & Beyond Homework  10  
[math analysis] sup f< sup g==>∫f^n<∫g^n  Calculus & Beyond Homework  3  
Solve 2+cos2x=3cosx  Calculus & Beyond Homework  1  
What does (cos(2x))^2 equal?  Precalculus Mathematics Homework  5 