What is the integral of sin^4 (x) dx?

Assume that, I = ∫ sin4x dx

Now, reduce sin4x dx to integrate the function.

I = ∫ (sin2x)2 dx

As we know 2sin2x = 1 – cos2x,

Put (sin2x)2 = {(1 − cos2x) / 2}2

= ∫ {(1 − cos2x) /2 }2 dx

= (1/4) ∫ (1 − cos2x)2 dx

As, (a – b)2 = a2 – 2ab + b2

= (1/4) ∫ (1− 2cos2x + cos22x) dx .(1)

Again reduce the function cos22x to integrate the function,

As, cos2(2x) = (1 + cos(4x)) / 2 … (2)

Now, substitute the value (2) in (1), we get

= (1/4) ∫ [1 – 2cos2x + {(1 + cos(4x)) / 2 }] dx

Now, integrate the function, we get

= (1/32)(12x – 8 sin (2x) + sin (4x) )+C

On simplifying the above expression, we get

= (3/8)x + (1/32) sin(4x) − (1/4) sin(2x) + C

Therefore, the integral of sin4x dx = (3/8)x + (1/32) sin(4x) − (1/4) sin(2x) + C

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question