Consider the integrals I = ∫ (sin x)/(3cos x + 4sin x)dx and J = ∫ (cos x)/(3cos x + 4sin x)dx. Then, the integral 4J - 3I is equal to (where, c is the constant of integration)

Correct answer: B. |3cos x + 4sin x| + c 4J - 3I = ∫ (4cos x - 3sin x)/(3cos x + 4sin x)dx Let, 3cos x + 4sin x = t ⇒ (4cos x - 3sin x)dx = dt ⇒ 4J - 3I = ∫ (dt)/(t) = |t| + c ⇒ 4J - 3I = |3cos x + 4sin x| + c

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Consider the integrals I = ∫ (sin x)/(3cos x + 4sin x)dx and J = ∫...

Consider the integrals

I = ∫ \dfrac{sin x}{3cos x + 4sin x}dx

and

J = ∫ \dfrac{cos x}{3cos x + 4sin x}dx.

Then, the integral

4J - 3I

is equal to (where,

c

is the constant of integration)

Options

A $\ln \left|3cos x - 4sin x\right| + c$
B $\ln \left|3cos x + 4sin x\right| + c$
C $\ln \left|3sin x + 4cos x\right| + c$
D $\ln \left|3sin x - 4cos x\right| + c$

Correct answer

B. $\ln \left|3cos x + 4sin x\right| + c$

Step-by-step solution

4J - 3I = ∫ \dfrac{4cos x - 3sin x}{3cos x + 4sin x}dx

Let,

3cos x + 4sin x = t

⇒ \left(4cos x - 3sin x\right)dx = dt

⇒ 4J - 3I = ∫ \dfrac{dt}{t} = \ln \left|t\right| + c

⇒ 4J - 3I = \ln \left|3cos x + 4sin x\right| + c

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