2x/² x + 3 x - 3 dx [AP EAPCET 21 May 2025 S1]

Correct Answer is : log⁡((cos⁡x−2)4(cos⁡x−1)2)+c≤ft(( x-2)^4/( x-1)²)+clog((cosx−1)2(cosx−2)4)+c

≤ft( ( x-2)²/≤ft| x-1| )+c

Correct Answer is : log⁡((cos⁡x−2)4(cos⁡x−1)2)+c≤ft(( x-2)^4/( x-1)²)+clog((cosx−1)2(cosx−2)4)+c

Test Index

Indefinite Integration

Section: Mathematics

Question : 36 of 89

Marks: +1, -0

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

Solution:

\;\text{ Let }\; I=\int \frac{\sin 2x}{\sin^2 x + 3\cos x - 3} dx

\;=-\int \frac{\sin 2x}{1-\cos^2 x+3\cos x-3} dx

\;=\int \frac{2 \sin x \cdot \cos x}{-\cos^2 x+3\cos x-2} dx

\;=\int \frac{2t(-dt)}{-t^2+3t-2} \qquad \text{Put } \cos x=t

\; \sin x dx = -dt

\;=\int \frac{2t dt}{t^2-3t+2} = \int \frac{2t}{(t-2)(t-1)} dt

\;=2 \int \left( \frac{2}{t-2} - \frac{1}{t-1} \right) dt

\;=2[2\ln(t-2)-\ln(t-1)]

\;=2\ln\left( \frac{(t-2)^2}{t-1} \right) = \ln \frac{(t-2)^4}{(t-1)^2} + C

\;=\ln\left(\frac{(\cos x-2)^4}{(\cos x-1)^2}\right)+C

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