When faced with the integral of 0.5 cos(x^2)^5, solving it using advanced techniques and trigonometric identities can be both challenging and rewarding. This guide will walk you through the process, breaking down each step to make it more comprehensible.

1. Initial Substitution and Identity Application

To start solving the integral, we initiate a substitution where u x^2. This transforms our original integral into 0.5 int cos(u^5) du.

1.1 Transforming the Integral

The integral to solve is:

I 0.5 int cos(u^5) du

1.2 Applying Trigonometric Identities

Next, we use a Pythagorean trigonometric identity: cos^2(u) sin^2(u) 1. By rearranging, we find:

cos^2(u) 1 - sin^2(u)

Squaring both sides, we get:

cos^4(u) (1 - sin^2(u))^2 1 - 2sin^2(u)sin^4(u)

2. Rewriting the Integrand and Substituting

Now, we rewrite the integrand using the trigonometric identity we derived:

0.5 int cos(u^5) du 0.5(1 - 2sin^2(u)sin^4(u)) du

Further, we separately write the integral as:

0.5 int (1 - 2cos(u)sin^2(u) - cos(u)sin^4(u)) du

3. Distributing and Integrating

Distribute the integral and integrate term by term:

0.5 int cos(u) du - 0.5(2 int cos(u)sin^2(u) du int cos(u)sin^4(u) du)

The first integral can be solved as:

0.5 sin(u)

3.1 Handling the Rest of the Integrals

We apply another substitution, letting w sin(u) so dw cos(u) du. This transforms our remaining integrals as follows:

-0.5 (2 int w^2 dw int w^4 dw)

Integrating these, we obtain:

-0.5(w^3/3 w^5/5)

Substituting back w sin(u) results in:

-0.5(sin^3(u)/3 sin^5(u)/5)

Combining all parts, the integral becomes:

0.5 sin(u) - 0.5(sin^3(u)/3 sin^5(u)/5)

Substituting u x^2 back in, we obtain our final solution:

0.5 sin(x^2) - 0.5(sin(x^2)^3/3 sin(x^2)^5/5)

This is a simplified and compact form of the solution.

4. Advanced Techniques and Product-to-Sum Formulas for Higher Powers

For more complex integrals, such as cos(x^2)^5, one can use product-to-sum formulas:

cos^5(u) Re{ (e^(iu))^5 } 1/16 Re{ e^(5iu) 5e^(3iu) 10e^(iu) }

Integrating each term separately using the integral formula int e^(au) du e^(au)/a, we get:

0.5 (1/16 (e^(5iu)/5 5e^(3iu)/3 10e^(iu)/1) C)

Transforming back into sine functions using Euler's formula, we have:

0.5 (1/16 (5 sin(5u)/5 5 sin(3u)/3 10 sin(u)/1) C)

Final form:

0.5 (sin(5u)/16 5 sin(3u)/48 10 sin(u)/16) C

Substituting u x^2 back in, we yield a sum of three sine functions.