INTEGRALS INVOLVING POWERS OF SINE AND COSINE

Problem 1 :

sin x cos5 x dx

Solution :

Given, sin x cos5 x dxcos x = t-sin x dx = dtsin x dx = -dtsin x cos5 x dx = t5 (-dt) = - t5 dt= -t5 + 15 + 1 + C= -t66 + C= -16 (cos x)6 + C= -16 cos6 x + C

Problem 2 :

sin3 x cos5 x dx

Solution :

Given, sin3 x cos5 x dx= sin2 x · sin x cos5 x dx= 1 - cos2 x sin x cos5 x dx= sin x cos5 x dx - sin x cos7 x dxcos x = t-sin x dx = dtsin x dx = -dtsin3 x cos5 x dx = t5 (-dt) - t7 (-dt) = - t5 dt + t7 dt= -t5 + 15 + 1 + t7 + 17 + 1 + C= -t66 + t88 + C=18 (cos x)8 - 16 (cos x)6 + C=18 cos8x - 16 cos6 x + C

Problem 3 :

Evaluate sin6 x cos5 x dx

Solution :

Given, sin6 x cos5 x dx= sin6 x cos4x · cos x dx=sin6 x 1 - sin2x2cos x dxsin x = tcos x dx = dtsin6 x cos5 x dx = t6 1 - t22 dt= t6 1 + t22 - 2(1)t2dt= t6 1 + t4 - 2t2dt= t6 + t10 - 2t8 dt= t6 dt + t10 dt - 2t8 dt= t77 + t1111 - 2t99 + C= 17sin7x + 111sin11x - 29sin9x + C

Problem 4 :

sin2 x cos2 x dx

Solution :

Given, sin2 x cos2 x dx=1 - cos 2x2 · 1 + cos 2x2 dx= 12 · 12 (1 - cos 2x) · (1 + cos 2x) dx= 14 1 - cos2 2x dx= 14dx - cos22x dx= 14x - 1 + cos 4x2dx= 14x - 12dx + cos 4xdx= 14x - 12x + 14 sin 4x + C= 14x - 12x - 18 sin 4x + C

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