Q90 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
The value of integral $\int x^3 \cos x\,dx$ is...
- A$x^3\sin x + x^2\cos x - 6x\sin x + 6\cos x + c$
- B$x^3\sin x + 3x^2\sin x - 6x\sin x - 6\cos x + c$
- Correct answerC$x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + c$
- D$x^3\sin x + 3x^2\cos x - 6x\sin x + 6\cos x + c$
Solution
Let $I = \int x^3 \cos x \, dx$.
Using integration by parts, $\int u \, dv = uv - \int v \, du$.
Taking $u = x^3$ and $dv = \cos x \, dx$, we get:
$I = x^3 \sin x - \int 3x^2 \sin x \, dx$
Applying integration by parts again for $\int 3x^2 \sin x \, dx$ with $u = 3x^2$ and $dv = \sin x \, dx$:
$I = x^3 \sin x - \left[ 3x^2 (-\cos x) - \int 6x (-\cos x) \, dx \right]$
$I = x^3 \sin x + 3x^2 \cos x - \int 6x \cos x \, dx$
Applying integration by parts once more for $\int 6x \cos x \, dx$ with $u = 6x$ and $dv = \cos x \, dx$:
$I = x^3 \sin x + 3x^2 \cos x - \left[ 6x \sin x - \int 6 \sin x \, dx \right]$
$I = x^3 \sin x + 3x^2 \cos x - 6x \sin x + \int 6 \sin x \, dx$
$I = x^3 \sin x + 3x^2 \cos x - 6x \sin x - 6 \cos x + c$
Answer: $x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + c$
Using integration by parts, $\int u \, dv = uv - \int v \, du$.
Taking $u = x^3$ and $dv = \cos x \, dx$, we get:
$I = x^3 \sin x - \int 3x^2 \sin x \, dx$
Applying integration by parts again for $\int 3x^2 \sin x \, dx$ with $u = 3x^2$ and $dv = \sin x \, dx$:
$I = x^3 \sin x - \left[ 3x^2 (-\cos x) - \int 6x (-\cos x) \, dx \right]$
$I = x^3 \sin x + 3x^2 \cos x - \int 6x \cos x \, dx$
Applying integration by parts once more for $\int 6x \cos x \, dx$ with $u = 6x$ and $dv = \cos x \, dx$:
$I = x^3 \sin x + 3x^2 \cos x - \left[ 6x \sin x - \int 6 \sin x \, dx \right]$
$I = x^3 \sin x + 3x^2 \cos x - 6x \sin x + \int 6 \sin x \, dx$
$I = x^3 \sin x + 3x^2 \cos x - 6x \sin x - 6 \cos x + c$
Answer: $x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + c$
Hence, the correct answer is option: 'C'.