Question

9 cos x - sin x
dx
5 cos x + 4 sin x​

Answer 1

Answer:

lf ∫

4sinx+5cosx

9cosx−sinx

d

X

=A Iog ∣4sinx+5cosx∣+ Bx +c, then A=………., B=…….,

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ANSWER

∫

4sin x+5 cos x

9 cos x−sin x

dx

N(x)=λ(D

1

(x))+μ(D(x))

⇒9 cos x−sin x=λ(4 cos x−5sin x)+μ(5 cos x+4 sin x)

4λ+5μ=9

−5λ+4mu=−1

} λ=μ=1

N(x)=D

1

(x)+D(x)

∫

D(x)

N(x)

dx= ∫

D(x)

D

1

(x)+D(x)

dx ∫

D(x)

D

1

(x)

+ ∫

D(x)

D(x)

dx

= log ∣D(x)∣+x+c

= log ∣4 sin x+5 cos x∣+x+c

A=1; B=1