The value of \(\int_{-1}^{1}\displaystyle (x^{3} + x\cos x + \tan^{5}x + 3)\,dx\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6
On a symmetric interval \([-a, a]\), the integral of an odd function is 0.
Each of \(x^{3}\), \(x\cos x\) and \(\tan^{5}x\) is odd, since replacing x by −x changes its sign:
\( (-x)^{3} = -x^{3} \)
\( (-x)\cos(-x) = -x\cos x \)
\( \tan^{5}(-x) = -\tan^{5}x \)
So only the constant term is left:
\( \displaystyle \int_{-1}^{1}3\,dx = 3\big(1 - (-1)\big) \)
\( = 6 \)