Question

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$(x3+xcosx+tan5x+1)dx is

• A. 0
• B. 2
• C. $\pi$
• D. 1

Answer 1

Answer: (C)

$\pi$

Answer 2

Let I=$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$(x3+xcosx+tan5x+1)dx

⇒I=$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$x3+dx+$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$cosx+$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$tan5xdx+$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$1. dx

It is known that if f(x)is an even function,then$\int_{-a}^{a}$ƒ(x)dx=2$\int_{0}^{a}$ƒ(x)dx and

if f(x)is an odd function,then∫a-aƒ(x)dx=0

I=0+0+0+2$\int_{0}^{\frac{\pi}{2}}$1.dx

=2[x]$^{\frac{\pi}{2}}_{2}$

=$\frac{2\pi}{2}$

=$\pi$

Hence,the correct Answer is C.