2 3x-3 3x/ 3x+2 3x\,dx = [TG EAPCET 9 May 2024 Shift 2]

Correct Answer is : 715log⁡∣cos⁡3x+2sin⁡3x∣−45x+c\;\; 7/15 3x+2 3x - 4/5 x +c157log∣cos3x+2sin3x∣−54x+c

Correct Answer is : 715log⁡∣cos⁡3x+2sin⁡3x∣−45x+c\;\; 7/15 3x+2 3x - 4/5 x +c157log∣cos3x+2sin3x∣−54x+c

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TS EAMCET 9 May 2024 Shift 2 Solved Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

\int \frac{2\cos 3x-3\sin 3x}{\cos 3x+2\sin 3x}\,dx =

Solution:

Let,

I = \int \frac{2\cos 3x-3\sin 3x}{\cos 3x+2\sin 3x}\,dx

Put

3x = t \Rightarrow 3\,dx = dt

\therefore \;\; I = \int \frac{2\cos t-3\sin t}{\cos t+2\sin t} \left(\frac{dt}{3}\right)

Let,

2\cos t-3\sin t = A(\cos t+2\sin t)

+ B \left[ \frac{d}{dt}(\cos t+2\sin t) \right]

\Rightarrow 2\cos t-3\sin t = A(\cos t+2\sin t)

\; + B(-\sin t+2\cos t)

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

On solving these equations, we get

\; A = -\frac{4}{3} \;\text{ and }\; B = \frac{7}{5}

\; \therefore I = \frac{1}{3} \left[ \int \frac{-\frac{4}{5}(\cos t+2\sin t)}{+\frac{7}{5}(-\sin t+2\cos t)}(\cos t+2\sin t] \,dt \right]

\; = \frac{1}{3} \left[ -\frac{4}{5} t + \frac{7}{5} \log \lvert \cos t+2\sin t \rvert \right] +C

\; = \frac{7}{15} \log \lvert \cos 3x+2\sin 3x \rvert - \frac{4}{5} x +C

\; \left[ \because t=3x \right]

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