#### To determine

**To evaluate:**

The given definite integral ∫-π/4π/4(x3+x4tanx)dx.

#### Answer

0

#### Explanation

**1) Concept:**

i) Integration of symmetric functions:

Suppose f is continuous on [-a,a]

a) If f is even f-x=fx then ∫-aafxdx=2∫0afxdx

b) If f is odd f-x=-fx then ∫-aafxdx=0

ii) tan-x=-tanx, tanx is odd function.

**2) Given:**

*∫-π/4π/4(x3+x4tanx)dx*

**3) Calculation:**

The given integral is ∫-π/4π/4(x3+x4tanx)dx Here,f(x)=(x3+x4tanx)

Check if given function is even or odd.

Replace x by –x in given f(x).

f-x=-x3+-x4tan-x

=-x3-x4tanx

=-x3+x4tanx

=-f(x)

Hence,f-x=-f(x)

The function is odd, therefore, by using the property of symmetric functions(odd function here) the solution of integral becomes 0.

∫-π/4π/4(x3+x4tanx)dx=0

**Conclusion:**

Therefore,

∫-π/4π/4(x3+x4tanx)dx=0