Q88 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics
The value of $\int \sin 4x \cos 3x\, dx$ is
- Correct answerA$-\dfrac{1}{14}\cos 7x - \dfrac{1}{2}\cos x + c$
- B$-\dfrac{1}{14}\cos 7x + \dfrac{1}{2}\cos x + c$
- C$\dfrac{1}{14}\cos 7x - \dfrac{1}{2}\cos x + c$
- D$\dfrac{1}{14}\cos 7x + \dfrac{1}{2}\cos x + c$
Solution
Let $I = \int \sin 4x \cos 3x \, dx$
Multiplying and dividing by $2$, we get:
$I = \dfrac{1}{2} \int 2 \sin 4x \cos 3x \, dx$
Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$ with $A = 4x$ and $B = 3x$:
$I = \dfrac{1}{2} \int (\sin 7x + \sin x) \, dx$
Integrating with respect to $x$:
$I = \dfrac{1}{2} \left( -\dfrac{\cos 7x}{7} - \cos x \right) + c$
$I = -\dfrac{1}{14} \cos 7x - \dfrac{1}{2} \cos x + c$
Answer: $-\dfrac{1}{14}\cos 7x - \dfrac{1}{2}\cos x + c$
Multiplying and dividing by $2$, we get:
$I = \dfrac{1}{2} \int 2 \sin 4x \cos 3x \, dx$
Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$ with $A = 4x$ and $B = 3x$:
$I = \dfrac{1}{2} \int (\sin 7x + \sin x) \, dx$
Integrating with respect to $x$:
$I = \dfrac{1}{2} \left( -\dfrac{\cos 7x}{7} - \cos x \right) + c$
$I = -\dfrac{1}{14} \cos 7x - \dfrac{1}{2} \cos x + c$
Answer: $-\dfrac{1}{14}\cos 7x - \dfrac{1}{2}\cos x + c$
Hence, the correct answer is option: 'A'.