∫ sin x ⋅ cos x ⋅ cos 2 x ⋅ cos 4 x ⋅ cos 8 x d x is equal to

Question

∫ sin x ⋅ cos x ⋅ cos 2 x ⋅ cos 4 x ⋅ cos 8 x d x is equal to (A) -(1/96) cos 16 x+C (B) -(1/256) cos 16 x+C (C) -(1/16) cos 16 x+C (D) None of these

Tardigrade Admin · Accepted Answer

I=∫ sin x ⋅ cos x ⋅ cos 2 x ⋅ cos 4 x ⋅ cos 8 x d x =(1/2) ∫ sin 2 x ⋅ cos 2 x ⋅ cos 4 x ⋅ cos 8 x d x =(1/4) ∫ sin 4 x ⋅ cos 4 x ⋅ cos 8 x d x =(1/8) ∫ sin 8 x ⋅ cos 8 x d x =(1/16) ∫ sin 16 x d x=(-1/256) cos 16 x+C

Q. $\int sin x \cdot cos x \cdot cos 2 x \cdot cos 4 x \cdot cos 8 x d x$ is equal to

$- \frac{1}{96} cos 16 x + C$

$- \frac{1}{256} cos 16 x + C$

$- \frac{1}{16} cos 16 x + C$

None of these

Q. ∫sinx⋅cosx⋅cos2x⋅cos4x⋅cos8xdx is equal to

−961​cos16x+C

−2561​cos16x+C

−161​cos16x+C

None of these

I=∫sinx⋅cosx⋅cos2x⋅cos4x⋅cos8xdx =21​∫sin2x⋅cos2x⋅cos4x⋅cos8xdx =41​∫sin4x⋅cos4x⋅cos8xdx =81​∫sin8x⋅cos8xdx =161​∫sin16xdx=256−1​cos16x+C

Q. $\int sin x \cdot cos x \cdot cos 2 x \cdot cos 4 x \cdot cos 8 x d x$ is equal to

$- \frac{1}{96} cos 16 x + C$

$- \frac{1}{256} cos 16 x + C$

$- \frac{1}{16} cos 16 x + C$