We can split cos³ x into cos² x · cos x. Using the fundamental trigonometric identity cos² x = 1 - sin² x, we express the integral entirely in terms of sin x accompanied by its derivative cos x.

Find Integral of cos³x dx

Answer: $\sin x - \frac{\sin^3 x}{3} + C$

Step-by-step solution

Step 1: Rewrite the integrand using trigonometric identity

We can split $\cos^3 x$ into $\cos^2 x \cdot \cos x$ . Using the fundamental trigonometric identity $\cos^2 x = 1 - \sin^2 x$ , we express the integral entirely in terms of $\sin x$ accompanied by its derivative $\cos x$ .

Step 2: Apply substitution method

We use the substitution method by setting $u = \sin x$ . Differentiating both sides with respect to $x$ gives $du = \cos x \, dx$ , which directly matches the differential element in our integrand.

Step 3: Substitute and integrate with respect to u

Substituting $u$ and $du$ transforms the integral into $\int (1 - u^2) \, du$ . Applying the power rule of integration, the integral of $1$ is $u$ , and the integral of $u^2$ is $\frac{u^3}{3}$ , where $C$ is the constant of integration.

Step 4: Substitute back for x

Finally, we substitute $u = \sin x$ back into our antiderivative to express the final result in terms of the original variable $x$ .

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