Question

The value of \(\int \frac{sin^3x}{(cos^4x+3cos^2x+1)tan^{-1}(secx+cosx)}\,dx\) is

• A. \(log(tan^{-1}(secx+cosx))+c\)
• B. \(2log(tan^{-1}(secx+cosx))+c\)
• C. \(\frac{(tan^{-1}(secx+cosx))^2}{2}+c\)
• D. \(tan^{-1}(secx+cosx)+c\)

Hint:

Substitute \(t=\cos x\) and notice the numerator is the derivative of \(\tan^{-1}(\sec x+\cos x)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The integrand has the form \(\dfrac{u'}{u}\) with \(u=\tan^{-1}(\sec x+\cos x)\). We show this by substitution.

Step 2: Key Formula or Approach
Let \(t=\cos x\), so \(\sin^3x\,dx=-(1-t^2)\,dt\). Let \(w=\sec x+\cos x=t+\dfrac1t\).

Step 3: Detailed Explanation
\(\dfrac{dw}{dt}=1-\dfrac{1}{t^2}=\dfrac{t^2-1}{t^2}\).
\[ \frac{d}{dt}\tan^{-1}w=\frac{(t^2-1)/t^2}{1+(t^2+1)^2/t^2}=\frac{t^2-1}{t^4+3t^2+1} \]
The integral becomes
\[ \int\frac{(t^2-1)\,dt}{(t^4+3t^2+1)\tan^{-1}w}=\int\frac{d(\tan^{-1}w)}{\tan^{-1}w}=\log\left(\tan^{-1}w\right)+c \]

Final Answer:
The integral is \(\log\left(\tan^{-1}(\sec x+\cos x)\right)+c\), option (A). \[ \boxed{\log\left(\tan^{-1}(\sec x+\cos x)\right)+c\ \text{(A)}} \]