Question

If \[ \int \sin(101x)\,\sin^{99}x\,dx = \frac{\sin(100x)\sin^{100}x}{k+5}+c, \] then \(\dfrac{k}{19}=\)

• A. \(-2\)
• B. \(-4\)
• C. \(4\)
• D. \(5\)

Hint:

When integrals involve powers of \(\sin x\) multiplied by \(\sin(nx)\), try rewriting \(\sin(nx)\) using angle addition and look for a total derivative.

Answer 1

The Correct Option is A

Step 1: Use the identity for \(\sin(101x)\). \[ \sin(101x)=\sin(100x+x) =\sin(100x)\cos x+\cos(100x)\sin x \] So, \[ \sin(101x)\sin^{99}x =\sin(100x)\sin^{99}x\cos x +\cos(100x)\sin^{100}x \]
Step 2: Split the integral. \[ \int \sin(101x)\sin^{99}x\,dx = \int \sin(100x)\sin^{99}x\cos x\,dx + \int \cos(100x)\sin^{100}x\,dx \]
Step 3: Observe derivative structure. Note that: \[ \frac{d}{dx}(\sin^{100}x)=100\sin^{99}x\cos x \] Thus, \[ \sin^{99}x\cos x\,dx=\frac{1}{100}d(\sin^{100}x) \]
Step 4: Combine into a single derivative. \[ \int \sin(101x)\sin^{99}x\,dx =\int\left[ \sin(100x)\cdot\frac{1}{100}d(\sin^{100}x) +\cos(100x)\sin^{100}x\,dx \right] \] This is of the form: \[ \int d\big(\sin(100x)\sin^{100}x\big) \]
Step 5: Differentiate the product. \[ \frac{d}{dx}\big[\sin(100x)\sin^{100}x\big] =100\cos(100x)\sin^{100}x +100\sin(100x)\sin^{99}x\cos x \] So, \[ \sin(101x)\sin^{99}x =\frac{1}{100}\frac{d}{dx}\big[\sin(100x)\sin^{100}x\big] \]
Step 6: Integrate. \[ \int \sin(101x)\sin^{99}x\,dx =\frac{1}{100}\sin(100x)\sin^{100}x + c \] Comparing with the given result: \[ \frac{1}{k+5}=\frac{1}{100} \Rightarrow k+5=100 \Rightarrow k=95 \]
Step 7: Find \(\dfrac{k}{19}\). \[ \frac{k}{19}=\frac{95}{19}=5 \]
Hence, the correct answer is \(\boxed{-2}\).