Prove that :\(\displaystyle\int_{-a}^{a}f(x)\,dx=2\int_0^af(x)\,dx\), if \(f\) is an even function\(=0\), if \(f\) is an odd function.Hence find the value of \(\displaystyle\int_{-1}^{1}\tan^{-1}x\,dx\).

Answer: Proved; \(\displaystyle\int_{-1}^1\tan^{-1}x\,dx=0\)

Proof: \(\displaystyle\int_{-a}^af(x)dx=\int_{-a}^0f(x)dx+\int_0^af(x)dx\) ...(1)

In \(\displaystyle\int_{-a}^0f(x)dx\) put \(x=-t\), \(dx=-dt\); when \(x=-a\), \(t=a\); when \(x=0\), \(t=0\).

\(\displaystyle\int_{-a}^0f(x)dx=\int_a^0f(-t)(-dt)=\int_0^af(-t)dt=\int_0^af(-x)dx\).

From (1): \(\displaystyle\int_{-a}^af(x)dx=\int_0^a[f(-x)+f(x)]dx\) ...(2)

If \(f\) is even, \(f(-x)=f(x)\), so (2) gives \(\displaystyle\int_{-a}^af(x)dx=2\int_0^af(x)dx\).

If \(f\) is odd, \(f(-x)=-f(x)\), so (2) gives \(\displaystyle\int_{-a}^af(x)dx=\int_0^a0\,dx=0\).

Hence part: Let \(f(x)=\tan^{-1}x\). Then \(f(-x)=\tan^{-1}(-x)=-\tan^{-1}x=-f(x)\), so \(f\) is odd. Therefore \(\displaystyle\int_{-1}^1\tan^{-1}x\,dx=0\).