Question

If \[ \int \frac{\sin^3x\left(\tan^{-1}(\sec x+\cos x)\right)^{-1}}{\cos^4x+3\cos^2x+1}\,dx=f(x)+C, \] then \[ e^{f(x)}= \]

• A. \(\tan^{-1}(\sec x+\cos x)\)
• B. \(\tan(\sec x+\cos x)\)
• C. \(\dfrac{1}{\cos^4x+3\cos^2x+1}\)
• D. \(\dfrac{\sin x}{\sin^3x+\cos^4x+1}\)

Hint:

If an integral has the form \[ \int \frac{u'}{u}\,dx, \] then its value is \[ \log u+C. \]

Answer 1

The Correct Option is A

Step 1: Observe the structure of the integrand.
The integral contains \[ \left(\tan^{-1}(\sec x+\cos x)\right)^{-1} \] So, we try to express the integrand in the form \[ \frac{u'}{u} \] Let \[ u=\tan^{-1}(\sec x+\cos x) \] Then, \[ f(x)=\log u \] will give \[ e^{f(x)}=u \]

Step 2: Differentiate \(u\).
Let \[ u=\tan^{-1}(\sec x+\cos x) \] Using \[ \frac{d}{dx}\tan^{-1}v=\frac{v'}{1+v^2}, \] where \[ v=\sec x+\cos x \] Now, \[ v'=\sec x\tan x-\sin x \] \[ =\frac{\sin x}{\cos^2x}-\sin x \] \[ =\sin x\left(\frac{1}{\cos^2x}-1\right) \] \[ =\sin x\left(\frac{1-\cos^2x}{\cos^2x}\right) \] \[ =\frac{\sin^3x}{\cos^2x} \] Also, \[ 1+v^2=1+(\sec x+\cos x)^2 \] \[ =1+\sec^2x+2+\cos^2x \] \[ =3+\frac{1}{\cos^2x}+\cos^2x \] \[ =\frac{3\cos^2x+1+\cos^4x}{\cos^2x} \] \[ =\frac{\cos^4x+3\cos^2x+1}{\cos^2x} \] Therefore, \[ u' = \frac{\frac{\sin^3x}{\cos^2x}} {\frac{\cos^4x+3\cos^2x+1}{\cos^2x}} \] \[ u'= \frac{\sin^3x}{\cos^4x+3\cos^2x+1} \]

Step 3: Rewrite the integral.
The given integral becomes \[ \int \frac{u'}{u}\,dx \] Therefore, \[ f(x)=\log u \] So, \[ f(x)=\log\left(\tan^{-1}(\sec x+\cos x)\right) \]

Step 4: Find \(e^{f(x)}\).
Taking exponential on both sides: \[ e^{f(x)} = e^{\log\left(\tan^{-1}(\sec x+\cos x)\right)} \] \[ e^{f(x)} = \tan^{-1}(\sec x+\cos x) \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\tan^{-1}(\sec x+\cos x)} \]