Question

Evaluate\(\int_\pi ^\pi {si{n^2}} xco{s^4}xdx\)

Short answer

The integral of\(\int_\pi ^\pi {si{n^2}} xco{s^4}xdx\)is \(0\).

Step-by-step solution

Concept of definite integral

if we reverse \(a\) and \(b\) in Theorem 4, then \(\Delta x\) changes from \((b - a)/n\) to \((a - b)/n\). Therefore

\(\int_b^a f (x)dx = - \int_a^b f (x)dx\)

If\(a = b\), then\(\Delta x = 0\)and so

\(\int_a^a f (x)dx = 0\)

Evaluate the integral

The integral is:

\(\int_\pi ^\pi {si{n^2}} xco{s^4}xdx\)

According to the concept,

If\(a = b\)then\(\Delta x = 0\)

And so\(\int_a^a f (x)dx = 0\)

Therefore,

The integral of \(\int_\pi ^\pi {si{n^2}} xco{s^4}xdx = 0\)