Integrals - Integration using trigonometric identities

Find $\int \cos^2 x \, dx$.

We use the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$. $\int \cos^2 x \, dx = \int \frac{1 + \cos 2x}{2} \, dx$ $= \frac{1}{2} \int (1 + \cos 2x) \, dx$ $= \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right] + C$ $= \frac{x}{2} + \frac{\sin 2x}{4} + C$

Direct integration of $\cos^2 x$ is not possible using basic rules, so we reduce the power from 2 to 1 using the double angle identity.

Evaluate $\int \sin 3x \cos 4x \, dx$.

Use the identity $2\sin A \cos B = \sin(A+B) + \sin(A-B)$. Here $A = 3x$ and $B = 4x$. $\int \sin 3x \cos 4x \, dx = \frac{1}{2} \int 2\sin 3x \cos 4x \, dx$ $= \frac{1}{2} \int [\sin(3x + 4x) + \sin(3x - 4x)] \, dx$ $= \frac{1}{2} \int [\sin 7x + \sin(-x)] \, dx$ Since $\sin(-x) = -\sin x$: $= \frac{1}{2} \int (\sin 7x - \sin x) \, dx$ $= \frac{1}{2} \left[ -\frac{\cos 7x}{7} - (-\cos x) \right] + C$ $= \frac{\cos x}{2} - \frac{\cos 7x}{14} + C$

When we have a product of sine and cosine with different angles, the product-to-sum identity transforms the product into a sum of two linear trigonometric functions.

Find $\int \sin^3 x \, dx$ using trigonometric identities.

We use the identity $\sin 3x = 3\sin x - 4\sin^3 x$, which gives $\sin^3 x = \frac{3\sin x - \sin 3x}{4}$. $\int \sin^3 x \, dx = \int \frac{3\sin x - \sin 3x}{4} \, dx$ $= \frac{1}{4} \left[ 3 \int \sin x \, dx - \int \sin 3x \, dx \right]$ $= \frac{1}{4} \left[ 3(-\cos x) - \left( -\frac{\cos 3x}{3} \right) \right] + C$ $= -\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C$

The triple angle identity is an efficient way to linearize powers of sine and cosine, making them immediately integrable.

We use the identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$. Let $A = 8x$ and $B = 4x$: $\int \sin 4x \sin 8x \, dx = \frac{1}{2} \int (2 \sin 8x \sin 4x) \, dx$ $\int \sin 4x \sin 8x \, dx = \frac{1}{2} \int [\cos(8x - 4x) - \cos(8x + 4x)] \, dx$ $\int \sin 4x \sin 8x \, dx = \frac{1}{2} \int (\cos 4x - \cos 12x) \, dx$ Integrating term by term: $\frac{1}{2} \left[ \frac{\sin 4x}{4} - \frac{\sin 12x}{12} \right] + C$ $= \frac{\sin 4x}{8} - \frac{\sin 12x}{24} + C$

To integrate the product of two sine functions with different frequencies, we transform the product into a difference of cosines using product-to-sum identities. This makes the integration straightforward as we only need to handle basic cosine terms.

We write $\cos^4 x = (\cos^2 x)^2$. Using $\cos^2 x = \frac{1 + \cos 2x}{2}$: $\cos^4 x = \left( \frac{1 + \cos 2x}{2} \right)^2 = \frac{1}{4} (1 + 2\cos 2x + \cos^2 2x)$ Now apply the identity to $\cos^2 2x$: $\cos^2 2x = \frac{1 + \cos 4x}{2}$ Substitute this back into the expression: $\cos^4 x = \frac{1}{4} \left( 1 + 2\cos 2x + \frac{1 + \cos 4x}{2} \right)$ $\cos^4 x = \frac{1}{4} \left( \frac{2 + 4\cos 2x + 1 + \cos 4x}{2} \right) = \frac{1}{8} (3 + 4\cos 2x + \cos 4x)$ Now integrate: $\int \cos^4 x \, dx = \frac{1}{8} \int (3 + 4\cos 2x + \cos 4x) \, dx$ $= \frac{1}{8} \left[ 3x + \frac{4\sin 2x}{2} + \frac{\sin 4x}{4} \right] + C$ $= \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$

Integrating $\cos^4 x$ requires reducing the degree from 4 to 1. This is achieved by applying the power reduction formula twice. The resulting expression is a sum of constant and basic cosine terms which are easily integrated.