Definite integral sin³x·cos²x from 0 to 1

This problem asks to evaluate the definite integral of the product of the cube of the sine and the square of the cosine on the interval [0, 1].

Problem

$\int_{0}^{1} {sin}^{3} x {cos}^{2} x d x$

Solution

Step 1

Note that sin³x = sin²x · sin x. This allows us to isolate one factor sin x, which can be brought under the differential.

Why do we isolate sin x specifically?

The integrand sin³x·cos²x contains an odd power of the sine (3) and an even power of the cosine (2). The standard technique: for an odd power of one function and any power of the other — isolate one factor of the function with the odd power and bring it under the differential. In our case, we isolate sin x, since d(cos x) = −sin x dx. The remaining sin²x is expressed in terms of cos²x using the basic trigonometric identity.

{sin}^{3} x = {sin}^{2} x \cdot sin x

Step 2

Rewrite the integral, isolating one factor sin x:

Why do we need the basic trigonometric identity?

After isolating one sin x, we are left with sin²x · cos²x · sin x dx. We replace the factor sin x dx with −d(cos x), but sin²x is still expressed in terms of sin x. To make the integral contain only cos x, we use the identity sin²x = 1 − cos²x. Then the entire expression becomes a function of cos x, and the integral is evaluated as a power integral.

\int_{0}^{1} {sin}^{2} x \cdot {cos}^{2} x sin x d x

Step 3

Apply the basic trigonometric identity sin²x = 1 − cos²x and note that sin x dx = −d(cos x):

{sin}^{2} x = 1 - {cos}^{2} x

sin x d x = - d (cos x)

Step 4

After the substitution t = cos x, the integral takes the form:

- \int_{0}^{1} (1 - {cos}^{2} x) {cos}^{2} x d (cos x)

Step 5

Expand the brackets:

- \int_{0}^{1} ({cos}^{4} x - {cos}^{2} x) d (cos x)

Step 6

Integrate as a power function and apply the Newton-Leibniz formula:

[_{0}^{1} \frac{{cos}^{3} x}{3} - \frac{{cos}^{5} x}{5}

Step 7

Substitute the limits. Note that cos 0 = 1, cos 1 ≈ 0.5403. We obtain the answer.

Answer

$\frac{2}{15}$

Frequently asked questions

How do I evaluate the definite integral sin³x·cos²x?

Isolate one factor sin x: sin³x = sin²x · sin x. Bring sin x under the differential: sin x dx = −d(cos x). Replace the remaining sin²x by 1 − cos²x. Then the integral reduces to a power integral in cos x.

Why do we isolate sin x and not cos x?

The power of the sine is odd (3), the power of the cosine is even (2). The standard rule: if one of the functions has an odd power, isolate one of its factors and bring it under the differential. In our case sin x dx = −d(cos x) is convenient because the remaining sin²x is easily expressed in terms of cos x. If the power of the cosine were odd, we would isolate cos x.

Why is the answer a rational number and not a trigonometric one?

Because after the substitution t = cos x we obtained an integral of a power function tⁿ, which is t^(n+1)/(n+1). The antiderivative is expressed in terms of powers of cos x, and substituting the limits (cos 0 = 1, cos 1 ≈ 0.5403) gives a numerical value. Often the answer contains fractional powers of cos 1, but in standard problems where the upper limit is π/2 or π, the answer is an integer or a rational number.

What if both powers are even?

If both powers are even, use the power reduction formulas: sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2. Apply them repeatedly until you obtain terms that can be integrated directly. This is the standard technique for integrals of the form ∫ sin^(2n)x cos^(2m)x dx.