\;2 x - 3 x/4 x - 3 x dx=

Correct Answer is : 125[log⁡∣4cos⁡x−3sin⁡x∣−18x]+C\;1/25[ |4 x -3 x|-18 x]+C251[log∣4cosx−3sinx∣−18x]+C

Correct Answer is : 125[log⁡∣4cos⁡x−3sin⁡x∣−18x]+C\;1/25[ |4 x -3 x|-18 x]+C251[log∣4cosx−3sinx∣−18x]+C

Test Index

TG EAMCET 2 May 2025 Shift 1 Paper

Question : 72 of 160

Marks: +1, -0

\int\;\frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx=

Solution:

I=\int\;\frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx

Let

t=4 \cos x-3 \sin x

\Rightarrow\;\frac{dt}{dx}=-4 \sin x-3 \cos x

Now,

2 \sin x-3 \cos x

\;\Rightarrow A(-4 \sin x-3 \cos x)+B(4 \cos x-3 \sin x)

\;\Rightarrow(-4A-3B) \sin x+(-3A+4B) \cos x

Comparing the coefficients, we get

-4A-3B=2\;\text{ and }\;-3A+4B=-3

Solving these two equations, we get

\;A=\;\frac{1}{25}\;\text{ and }\;B=\;\frac{-18}{25}

\;\;\text{ So, }\; I=\int\;\frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx

\;\Rightarrow\;\frac{1}{25}\int\;\frac{-4 \sin x -3 \cos x}{4 \cos x -3 \sin x} dx+ \left(\;\frac{-18}{25}\right) \int\;\frac{4 \cos x -3 \sin x}{4 \cos x -3 \sin x} dx

\;\Rightarrow\;\frac{1}{25}\int\;\frac{dt}{t}-\;\frac{18}{25}\int dx \Rightarrow\;\frac{1}{25}\log |t|-\;\frac{18}{25} x+C

\;\Rightarrow\;\frac{1}{25}\log |4 \cos x -3 \sin x|-\;\frac{18}{25} x+C

\;=\;\frac{1}{25}[\log |4 \cos x -3 \sin x|-18 x]+C

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