#### To determine

**To evaluate:**

∫-π/4π/4t4 tant2+cost dt if it exists

#### Answer

0

#### Explanation

**1) Concept:**

By using theorem for integration of an odd function

**2) Theorem:**

Suppose f is continuous on [-a, a] and if f is odd [f-x=-fx], then

∫-aaf(x)dx=0

**3) Given:**

∫-π/4π/4t4 tant2+cost dt

**4) Calculation:**

Consider,

∫-π/4π/4t4 tant2+cost dt

Here,

ft=t4 tant2+cost

Now,

f-t=(-t)4tan(-t)2+cos(-t)

Since tan-t=-tant and cos(-t) = cos(t)

f-t=-t4 tant2+cost=-f(t)

Therefore, the function is odd; also it is continuous in thegiven interval, so, its integral exists.

From the theorem

∫-π/4π/4t4 tant2+cost dt=0

**Conclusion:**

Therefore,

∫-π/4π/4t4 tant2+cost dt=0