#### To determine

**To evaluate:**

The given indefinite integral ∫sinxcos4x dx.

#### Answer

∫sinxcos4x dx=-15(cos5x)+C

From the graph, it is reasonable to say -15(cos5x)+C is the antiderivative of sinxcos4x.

#### Explanation

**1) Concept:**

i) The substitution rule

If u=g(x) is a differentiable function whose range is I and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du. ii) Indefinite integral

∫xn dx=xn+1n+1+C n≠-1

**2) Given:**

∫sin3xcosx dx

**3) Calculation:**

The given integral is ∫sinxcos4x dx.

Here, use the substitution method.

Substitute cosx=u.

Differentiating with respect to x,

-sinx dx=du

sinx dx=-du

Therefore, the given integral becomes

∫sinxcos4x dx=∫u4(-du)

=-∫u4du

By using concept ii),

=-u4+14+1+C

=-u55+C

Resubstitute sinx=u.

Hence, the solution becomes

=-15(cos5x)+C

Now, take C=0 and sketch the graph of the function fx= sinxcos4x and its antiderivative Fx=-15(cos5x).

The blue graph is for Fx=-15(cos5x) and the red graph is for fx= sinxcos4x.

Where f(x) is positive, there Fx is increasing; where f(x) is negative, there Fx is decreasing;where f(x) changes from negative to positive, there Fx has a local minimum; and where f(x) changes from positive to negative, there Fx has a local maximum. Both functions f(x) and F(x) are periodic with period 2π. So it is reasonable to say that F is anti-derivative of f.

**Conclusion:**

Therefore,

∫sinxcos4x dx=-15(cos5x)+C

From the graph, it is reasonable to say -15(cos5x)+C is the antiderivative of sinxcos4x.