Integral Calculation - Exercise 2

We aim to evaluate the following indefinite integral:

$$ \int \sin^3(x) \cdot \cos(x) \ dx $$

We'll approach this using the substitution method.

Let’s set $ t = \sin(x) $ as our substitution:

$$ t = \sin(x) $$

We now differentiate both sides, treating the left-hand side as a function of $ t $ and the right-hand side as a function of $ x $:

$$ dt = \cos(x) \ dx $$

Substituting $ \sin(x) = t $, we rewrite $ \sin^3(x) $ as $ t^3 $:

$$ \int t^3 \cdot \cos(x) \ dx $$

Since $ dt = \cos(x) \ dx $, the integral becomes:

$$ \int t^3 \ dt $$

This is now a straightforward power function.

Recall that the antiderivative of $ t^n $ is given by:

$$ \int t^n \ dt = \frac{t^{n+1}}{n+1} + c $$

Applying the rule for $ n = 3 $:

$$ \int t^3 \ dt = \frac{t^4}{4} + c $$

Substituting back $ t = \sin(x) $, we obtain the final result:

$$ \frac{\sin^4(x)}{4} + c $$

Alternative approach. This integral $$ \int \sin^3(x) \cdot \cos(x) \ dx $$ can also be solved using a different integration technique: $$ \int f'(x) \cdot [f(x)]^n \ dx = \frac{[f(x)]^{n+1}}{n+1} + c $$ Here, we take $ f(x) = \sin(x) $, $ f'(x) = \cos(x) $, and $ n = 3 $. Applying the formula gives: $$ \int \cos(x) \cdot [\sin(x)]^3 \ dx = \frac{\sin^4(x)}{4} + c $$ As expected, we arrive at the same result.

And so on...