integral of xcos2xsin4x
Answers
Answer:
Step-by-step explanation:
∫cos 2x sin 4x dx
∫(2cos ^2 x -1)sin 4x dx
∫2cos^2 x*sin 4x dx -∫sin 4x dx
2∫cos^2 x*2sin2xcos 2x dx -∫sin 4x dx
2∫cos ^2 x*4sinxcosx(2cos^2 x -1) dx -∫ sin 4x dx
8∫sinxcos^3 x *(2cos^2 x -1) dx -∫sin 4x dx
8{ ∫2sinxcos^5 x dx - ∫sinxcos^3 x dx } -∫sin 4x dx
8{2∫sinxcos^5 x dx - ∫sinxcos^3 x dx } +cos 4x /4
assume t as cos x in above
8{-2∫t^5 dt + ∫t^3 dt } + cos 4x /4
8{-2t^6/6+t^4/4 } + cos 4x/4
8{-t^6/3+t^4/4}+cos 4x/4
-8/3 cos^6 x +cos^4 x /4 +cos 4x/4
therefore integral xcos2xsin4x
applying integration by parts
x∫cos2xsin4x dx -∫ (-8/3 cos^6 x +cos^4 x /4 +cos 4x/4) dx
=x(-8/3 cos^6 x +cos^4 x /4 +cos 4x/4 ) - ∫ (-8/3 cos^6 x +cos^4 x /4 +cos 4x/4) dx
=x(-8/3 cos^6 x +cos^4 x /4 +cos 4x/4 ) + 8/3∫cos^6 x -1/4∫cos^4 x dx
-1/4 ∫cos 4x dx
=x(-8/3 cos^6 x +cos^4 x /4 +cos 4x/4 ) + 8/3 (cos^5 xsinx /6 +5/6(-cos^4 x /16 + 3/32sin 2x +3/16 x^2) -1/4{cos^3 xsinx/4+3/16 cos 2x + 3x/8 } -1/4 (sin4x/4)