Given,cos2(2x) dx=1 + cos (2 × 2x)2 dx=1 + cos 4x2 dx=12 1 + cos 4x dx=12dx +cos 4x dx=12x +sin 4x4+ C=12x +18 sin 4x + C
Problem 3 :
cos3 x dx
Solution :
Given,cos3 x dx=cos2x cos x dx=1 - sin2x cos x dxu = sin x, du = cos x dx=1 - u2 du=1 du -u2 du= u -u33+ C= sin x -13sin3x + C
Problem 4 :
cos4 x dx
Solution :
Given,cos4 x dx=cos2 x2 dx=1 + cos 2x22 dx=(1 + cos 2x)222 dx=(1 + cos 2x)24 dx=14(1 + cos 2x)2 dx=1412+(cos 2x)2 + 2(1)(cos 2x) dx=141 +cos2 2x + 2 cos 2x dx=14dx +cos2 2x dx +2 cos 2x dx=14x +x2+sin 4x8+ sin 2x+ C=14x +18x +132sin 4x +14sin 2x +C=38x +132sin 4x +14sin 2x +C
Problem 5 :
cos5 x dx
Solution :
Given,cos5 x dx=cos4x cos x dx=1 - sin2x2 cos x dxu = sin x, du = cos x dx=1 - u22 du=12+u22- 2(1)u2 du=1 + u4- 2u2 du=1 du +u4 du -2u2 du= u +u55-2u33+ C= sin x +15sin5x -23sin3x + C