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Use Wallis's formulas to evaluate the integral ∫(20sin⁹(x)) dx.

User Vitalijs
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Final answer:

To evaluate the integral ∫(20sin⁹(x)) dx using Wallis's formulas, apply a trigonometric identity and a substitution. The result is (-5/4)cos⁴(x) + C.

Step-by-step explanation:

To evaluate the integral ∫(20sin⁹(x)) dx using Wallis's formulas, we need to apply a trigonometric identity and a substitution. First, let's rewrite the integral using the identity sin⁹(x) = (sin(x))⁹. Now, let's substitute u = cos(x) so that du = -sin(x) dx. The integral becomes -20 ∫(u⁹) du.

Now, we can use the power rule for integrals to evaluate -20 ∫(u⁹) du. Applying the power rule, we get (-20/(4))u⁺ + C, where C is the constant of integration. Finally, substituting u back as cos(x), the result is (-5/4)cos⁺(x) + C.

User Mslowiak
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