The value of 4 x 3 x d x is

Correct Answer is : −114cos⁡7x−12cos⁡x+c-1/14 7 x-1/2 x+c−141cos7x−21cosx+c

Correct Answer is : −114cos⁡7x−12cos⁡x+c-1/14 7 x-1/2 x+c−141cos7x−21cosx+c

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MHTCET 20 Apr 2026 Shift 1 Paper

Question : 130 of 150

Marks: +1, -0

The value of

\int \sin 4 x \cos 3 x d x

Solution:

Concept:

Use the trigonometric identity

\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]

to simplify the product before integrating.

Explanation:

Apply the identity to

\sin 4x \cos 3x

Here,

A = 4x

and

B = 3x

So,

\sin 4x \cos 3x = \frac{1}{2}[\sin(4x+3x) + \sin(4x-3x)]

This simplifies to

\frac{1}{2}[\sin 7x + \sin x]

Now integrate term by term.

\int \sin 4x \cos 3x \, dx = \int \frac{1}{2}(\sin 7x + \sin x) \, dx

= \frac{1}{2}\left(-\frac{\cos 7x}{7} - \cos x\right) + c

= -\frac{1}{14}\cos 7x - \frac{1}{2}\cos x + c

This matches option A exactly.

Answer:

Option A:

-\frac{1}{14} \cos 7x - \frac{1}{2} \cos x + c

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