Evaluating the Integral of sin3 x · 1/2 cos x · cos x2 over [0, π]

In this article, we will walk through the detailed steps to evaluate the following integral:

( I int_{0}^{pi} sin^3 x cdot frac{1}{2} cos x cos x^2 , dx )

Step 1: Expansion of the Integrand

The integrand in the given integral is ( frac{1}{2} cos x cos x^2 ). We start by expanding this expression:

( 1 - cos x^2 1 - frac{1 cos 2x}{2} frac{1 - cos 2x}{2} )

Using the product-to-sum identities, we get:

( frac{1}{2} cos x (1 - cos x^2) frac{1}{2} cos x left( frac{1 - cos 2x}{2} right) frac{1}{4} cos x - frac{1}{4} cos x cos 2x )

Next, expanding ( cos x cos 2x ) using product-to-sum identities:

( cos x cos 2x frac{1}{2} (cos 3x cos x) )

Thus, the expression becomes:

( frac{1}{4} cos x - frac{1}{4} left( frac{1}{2} (cos 3x cos x) right) frac{1}{4} cos x - frac{1}{8} cos 3x - frac{1}{8} cos x frac{1}{8} cos x - frac{1}{8} cos 3x )

Step 2: Substitution Back into the Integral

Substituting the expanded form back into the original integral:

( I int_{0}^{pi} sin^3 x left( frac{1}{8} cos x - frac{1}{8} cos 3x right) dx )

We split this into two separate integrals:

( I frac{1}{8} int_{0}^{pi} sin^3 x cos x , dx - frac{1}{8} int_{0}^{pi} sin^3 x cos 3x , dx )

Step 3: Evaluating Each Integral

First Integral:

( int_{0}^{pi} sin^3 x cos x , dx )

We use the substitution ( u sin x ), ( du cos x , dx ). The limits change from ( 0 ) to ( pi ) to ( 0 ) to ( 0 ), which makes the integral evaluate to zero.

Second Integral:

( int_{0}^{pi} sin^3 x cos 3x , dx )

Using the product-to-sum identities, we get:

( sin^3 x frac{3 sin x - sin 3x}{4} )

Then the integral becomes:

( int_{0}^{pi} left( frac{3 sin x - sin 3x}{4} right) cos 3x , dx frac{1}{4} int_{0}^{pi} (3 sin x cos 3x - sin 3x cos 3x) , dx )

This simplifies to:

( frac{1}{4} int_{0}^{pi} 3 sin x cos 3x , dx - frac{1}{4} int_{0}^{pi} sin^2 3x , dx )

The first integral evaluates to zero due to symmetry, and the second integral evaluates to ( frac{pi}{6} ).

Combining Results:

( I 0 - frac{1}{8} cdot frac{pi}{6} -frac{pi}{48} )

Conclusion

The final value of the integral is:

( I -frac{pi}{48} )

By carefully expanding the integrand, applying substitution, and simplifying using trigonometric identities, we have successfully evaluated the given integral.