1/ x ≤ft[ 1/ x - 1/ x+3 x ] dx = [AP EAPCET 23 May 2025 S1]

Correct Answer is : log⁡∣sin⁡xsin⁡x+3cos⁡x∣+c ≤ft| x/ x+3 x | + clogsinx+3cosxsinx+c

Correct Answer is : log⁡∣sin⁡xsin⁡x+3cos⁡x∣+c ≤ft| x/ x+3 x | + clogsinx+3cosxsinx+c

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Indefinite Integration

Section: Mathematics

Question : 9 of 89

Marks: +1, -0

\int \frac{1}{\cos x} \left[ \frac{1}{\sin x} - \frac{1}{\sin x+3\cos x} \right] dx =

Solution:

Let

I = \int \frac{1}{\cos x} \left[ \frac{1}{\sin x} - \frac{1}{\sin x+3\cos x} \right] dx

= \int \frac{2}{\sin 2x} dx - \int \frac{\sec^2 x}{\tan x+3} dx

= 2 \int \csc 2x \, dx - \int \frac{dz}{z+3}, \quad \text{ where }

z = \tan x

= \frac{2 \log_e |\csc 2x - \cot 2x|}{2} - \ln |z+3| + C

= \ln |\csc 2x - \cot 2x| - \ln |\tan x+3| + C

= \ln \left| \frac{ \frac{1}{\sin 2x} - \frac{\cos 2x}{\sin 2x} }{ \frac{\sin x}{\cos x} + 3 } \right| + C

= \ln \left| \frac{ \frac{1-\cos 2x}{\sin 2x} }{ \frac{\sin x+3\cos x}{\cos x} } \right| + C

= \ln \left| \frac{2 \sin 2}{2 \sin x \cdot \cos x} \times \frac{\cos x}{\sin x+3\cos x} \right| + C

= \ln \left| \frac{\sin x}{\sin x+3\cos x} \right| + C

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