If ∫ sin 3 x cos 5 x d x=A sin 4 x +B sin 6 x +C sin 8 x+ D. Then

Question

If ∫ sin 3 x cos 5 x d x=A sin 4 x +B sin 6 x +C sin 8 x+ D. Then (A) A =(1/4), B =-(1/3), C =(1/8), D ∈ R (B) A =(1/8), B =(1/4), C =(1/3), D ∈ R (C) A =0, B =-(1/6), C =(1/8), D ∈ R (D) None of these

Tardigrade Admin · Accepted Answer

I =∫ sin 3 x ⋅ cos 5 x dx Put sin x=t ⇒ cos x d x=d t I =∫ sin 3 x ⋅ cos 4 x ⋅ cos x d x =∫ t 3(1- t 2)2 dt =∫(t3-2 t5+t7) d t=(1/4) t4-(2/6) t6+(1/8) t8+D =(1/4) sin 4 x-(1/3) sin 6 x+(1/8) sin 8 x+D

Q. If $\int sin^{3} x cos^{5} x d x = A sin^{4} x + B sin^{6} x + C sin^{8} x + D$ . Then

$A = \frac{1}{4}, B = - \frac{1}{3}, C = \frac{1}{8}, D \in R$

$A = \frac{1}{8}, B = \frac{1}{4}, C = \frac{1}{3}, D \in R$

$A = 0, B = - \frac{1}{6}, C = \frac{1}{8}, D \in R$

None of these

Q. If ∫sin3xcos5xdx=Asin4x+Bsin6x+Csin8x+D. Then

A=41​,B=−31​,C=81​,D∈R

A=81​,B=41​,C=31​,D∈R

A=0,B=−61​,C=81​,D∈R

None of these

I=∫sin3x⋅cos5xdx Put sinx=t⇒cosxdx=dt I=∫sin3x⋅cos4x⋅cosxdx=∫t3(1−t2)2dt =∫(t3−2t5+t7)dt=41​t4−62​t6+81​t8+D =41​sin4x−31​sin6x+81​sin8x+D

Q. If $\int sin^{3} x cos^{5} x d x = A sin^{4} x + B sin^{6} x + C sin^{8} x + D$ . Then

$A = \frac{1}{4}, B = - \frac{1}{3}, C = \frac{1}{8}, D \in R$

$A = \frac{1}{8}, B = \frac{1}{4}, C = \frac{1}{3}, D \in R$

$A = 0, B = - \frac{1}{6}, C = \frac{1}{8}, D \in R$