JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int\limits_{{-1}}^1 x^2 \sin 2 x d x=

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Correct option: B

-\frac{1}{2} x^2 \cos 2 x+\frac{1}{2} x \sin 2 x+\frac{1}{4} \cos 2 x+c

Option B,

-\frac{1}{2} x^2 \cos 2 x+\frac{1}{2} x \sin 2 x+\frac{1}{4} \cos 2 x+c

, is the correct antiderivative, but the question asks for the definite integral from -1 to 1 of an odd function, which is 0.

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