\intx^2\sinx\cosx\dx=

Correct Answer is : (1−2x2)82cos2x+x4sin2x+c

-\frac{x^2\cos\2x}{2}+{x\sin2x}/4+{\cos\2x}/8+c

{(1-2x^2)}/82cos\2x+x/4\sin\2x+c

{(1-2x^2)^2}/82cos\2x+x/2\sin\2x+c

AP EAMCET Engineering 25 Apr 2017 Shift 1 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

∫x2‌sinx‌cosx‌dx=

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