Question

Evaluate the integrals in Exercises 1–22.∫₀^π sin⁵(x/2) dx

Accepted Answer

Recognize that the integral is $$ \int_0^{\pi} \sin^5\left(\frac{x}{2}\right) \, dx . $$The integrand is a sine function raised to an odd power, which suggests using a reduction formula or rewriting the power in terms of lower powers of sine and cosine. Use the substitution $$ u = \frac{x}{2} $$, which implies $$ du = \frac{1}{2} dx  or $$equivalently $$ dx = 2 du . $$Also, change the limits of integration accordingly: when $$ x = 0 $$, $$ u = 0 $$, and when $$ x = \pi $$, $$ u = \frac{\pi}{2} . $$Rewrite the integral in terms of $$ u $$: $$ \int_0^{\pi} \sin^5\left(\frac{x}{2}\right) dx = \int_0^{\frac{\pi}{2}} \sin^5(u) \cdot 2 \, du = 2 \int_0^{\frac{\pi}{2}} \sin^5(u) \, du. $$ Express $$ \sin^5(u)  as$$ $$ \sin^4(u) \sin(u) $$ and use the identity $$ \sin^2(u) = 1 - \cos^2(u)  to $$rewrite $$ \sin^4(u) = (\sin^2(u))^2 = (1 - \cos^2(u))^2 . $$This allows the integral to be expressed as $$ 2 \int_0^{\frac{\pi}{2}} (1 - \cos^2(u))^2 \sin(u) \, du. $$ Use the substitution $$ t = \cos(u) $$, so $$ dt = -\sin(u) du  or$$ $$ -dt = \sin(u) du . $$Change the limits for $$ t $$: when $$ u = 0 $$, $$ t = \cos(0) = 1 $$, and when $$ u = \frac{\pi}{2} $$, $$ t = \cos\left(\frac{\pi}{2}\right) = 0 . $$Rewrite the integral in terms of $$ t $$ and integrate the resulting polynomial expression.

Evaluate the integrals in Exercises 1–22.
∫₀^π sin⁵(x/2) dx

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Key Concepts

Definite Integrals

Trigonometric Integrals

Substitution Method