\int \frac{1}{9 \cos ^2 x-24 \sin x \cos x+16 \sin ^2 x} \dx= [AP EAPCET 22 May 2025 S1]

Correct Answer is : cosx4(3cosx−4sinx)+c

\frac{\cos x}{4(3 \cos x-4 \sin x)}+c

\frac{ \sin x}{4(3 \cos x-4 \sin x)}+c

\frac{\cos x}{3 \cos x-4 \sin x}+c

\frac{ \sin x}{3 \cos x-4 \sin x}+c

Indefinite Integration

Section: Mathematics

Question : 15 of 89

Marks: +1, -0

∫‌

9cos2x−24sin‌x‌cos‌x+16sin‌2x

‌dx=

[AP EAPCET 22 May 2025 S1]

Go to Question:

Prev Question

Next Question