\int \sin ^4 x \cos ^4 x \dx= [AP EAPCET 22 May 2024 Shift 2]

Correct Answer is : 1256(−2sin32xcos2x−3sin2xcos2x+6x)+c

\frac{1}{128} (-2 \sin ^3 x \cos x-3 \sin x \cos x+3) +c

\frac{1}{256} (-2 \sin ^3 2 x \cos 2 x-3 \sin 2 x \cos 2 x+6 x) +c

\frac{1}{128} (2 \sin ^3 x \cos x-3 \sin x \cos x+3 x) +c

\frac{1}{256} (3 \sin ^3 x \cos x-2 \sin x \cos x+2) +c

AP EAPCET 22 May 2024 Shift 2 Solved Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

∫sin‌4xcos4x‌dx=

[AP EAPCET 22 May 2024 Shift 2]

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