Question

If \[ f(x)=\int x^2\cos^2x\left(2x\tan^2x-2x-6\tan x\right)\,dx \] and \(f(0)=\pi\), then \(f(x)=\)

• A. \(x^2\sin x+\pi\)
• B. \(\cos x+\pi-1\)
• C. \(-x^3\sin 2x+\pi\)
• D. \(x^3\cos 2x+\pi\cos x\)

Hint:

When an integrand contains \(\tan x\) and \(\cos^2x\), first rewrite \(\tan x\) in terms of \(\sin x\) and \(\cos x\), then use identities like \(\sin 2x=2\sin x\cos x\) and \(\cos 2x=\cos^2x-\sin^2x\).

Answer 1

The Correct Option is C

Step 1: Simplify the integrand.
Given, \[ f(x)=\int x^2\cos^2x\left(2x\tan^2x-2x-6\tan x\right)\,dx \] Using \[ \tan x=\frac{\sin x}{\cos x} \] and \[ \tan^2x=\frac{\sin^2x}{\cos^2x}, \] we get \[ x^2\cos^2x\left(2x\tan^2x-2x-6\tan x\right) \] \[ = x^2\cos^2x \left( 2x\frac{\sin^2x}{\cos^2x} -2x -6\frac{\sin x}{\cos x} \right) \] \[ = x^2\left(2x\sin^2x-2x\cos^2x-6\sin x\cos x\right) \]

Step 2: Use trigonometric identities.
We know that \[ \cos 2x=\cos^2x-\sin^2x \] So, \[ \sin^2x-\cos^2x=-\cos 2x \] Also, \[ 2\sin x\cos x=\sin 2x \] Therefore, \[ 2x\sin^2x-2x\cos^2x = 2x(\sin^2x-\cos^2x) \] \[ = -2x\cos 2x \] and \[ -6\sin x\cos x=-3\sin 2x \] Hence, the integrand becomes \[ x^2\left(-2x\cos 2x-3\sin 2x\right) \] \[ = -2x^3\cos 2x-3x^2\sin 2x \]

Step 3: Identify the antiderivative.
Now observe that \[ \frac{d}{dx}\left(-x^3\sin 2x\right) \] Using product rule, \[ = -\left(3x^2\sin 2x+x^3\cdot 2\cos 2x\right) \] \[ = -3x^2\sin 2x-2x^3\cos 2x \] This is exactly the simplified integrand.
Therefore, \[ f(x)=-x^3\sin 2x+C \]

Step 4: Use the given condition \(f(0)=\pi\).
Substitute \(x=0\): \[ f(0)=-(0)^3\sin 0+C \] \[ f(0)=C \] Given that \[ f(0)=\pi \] So, \[ C=\pi \] Therefore, \[ f(x)=-x^3\sin 2x+\pi \]

Step 5: Final conclusion.
Hence, \[ \boxed{-x^3\sin 2x+\pi} \]