Definite integral cos³x·sin⁴x from 0 to π/2

This problem asks to evaluate the definite integral of the product of the cube of the cosine and the fourth power of the sine on the interval [0, π/2].

Problem

$\int_{0}^{π / 2} {cos}^{3} x {sin}^{4} x d x$

Solution

Step 1

Note that cos³x = cos²x · cos x. This allows us to isolate one factor cos x, which can be brought under the differential.

Why do we isolate cos x specifically?

The integrand cos³x·sin⁴x contains an odd power of the cosine (3) and an even power of the sine (4). 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 cos x, since d(sin x) = cos x dx. The remaining cos²x is expressed in terms of sin²x using the basic trigonometric identity.

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

Step 2

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

Why do we need the basic trigonometric identity?

After isolating one cos x, we are left with cos²x · sin⁴x · cos x dx. We replace the factor cos x dx with d(sin x), but cos²x is still expressed in terms of cos x. To make the integral contain only sin x, we use the identity cos²x = 1 − sin²x. Then the entire expression becomes a function of sin x, and the integral is evaluated as a power integral.

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

cos x d x = d (sin x)

Step 3

Rewrite the integral, substituting the identity and replacing cos x dx with d(sin x):

\int_{0}^{π / 2} (1 - {sin}^{2} x) {sin}^{4} x d (sin x)

Step 4

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

\int_{0}^{1} (t^{4} - t^{6}) d t

Step 5

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

[_{0}^{1} \frac{t^{5}}{5} - \frac{t^{7}}{7}

Step 6

Substitute the limits t = 1 and t = 0. We obtain the answer.

Answer

$\frac{2}{35}$

Frequently asked questions

How do I evaluate the definite integral cos³x·sin⁴x?

Isolate one factor cos x: cos³x = cos²x · cos x. Bring cos x under the differential: cos x dx = d(sin x). Replace the remaining cos²x by 1 − sin²x. Then the integral reduces to a power integral in sin x.

Why do we isolate cos x and not sin x?

The power of the cosine is odd (3), the power of the sine is even (4). 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 cos x dx = d(sin x) is convenient because the remaining cos²x is easily expressed in terms of sin x. If the power of the sine were odd, we would isolate sin x.

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

Because after the substitution t = sin x we obtained an integral of a power function tⁿ, which is tⁿ⁺¹/(n + 1). The antiderivative is expressed in terms of powers of sin x, and substituting the limits (sin 0 = 0, sin(π/2) = 1) gives a numerical value.

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.