If $\int \sin 5 x \cos 3 x d x=-\frac{\cos 8 x}{16}+A$, then $A=$

$\frac{\sin 2 x}{16}+$ constant

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

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Application +4 | -1

\int \sin 5 x \cos 3 x d x=-\frac{\cos 8 x}{16}+A

, then

A=

View answer and explanation

Correct option: B

-\frac{\cos 2 x}{4}+

constant

Option B,

-\frac{\cos 2 x}{4}+

constant, is the correct answer because it is derived from integrating the product of sine and cosine functions using trigonometric identities.

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