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Find: integral of sin³(x)cos(x/2)

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Find: integral of sin³(x)cos(x/2)
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Manav Sharma 1 month, 2 weeks ago

To find the integral of \( \sin^3(x) \cos\left(\frac{x}{2}\right) \), we can use a substitution. Let: \[ u = \sin(x) \] \[ du = \cos(x) dx \] Now, we can rewrite the integral as: \[ \int u^3 \cos\left(\frac{x}{2}\right) du \] Since \( du = \cos(x) dx \), we need to express \( \cos(x) \) in terms of \( u \). From the trigonometric identity \( \sin^2(x) + \cos^2(x) = 1 \), we have: \[ \cos^2(x) = 1 - \sin^2(x) \] \[ \cos(x) = \sqrt{1 - u^2} \] Now, substitute \( \cos(x) = \sqrt{1 - u^2} \) into the integral: \[ \int u^3 \sqrt{1 - u^2} du \] This integral can be solved using trigonometric substitution. Let: \[ u = \sin(\theta) \] \[ du = \cos(\theta) d\theta \] Now, rewrite the integral in terms of \( \theta \): \[ \int \sin^3(\theta) \sqrt{1 - \sin^2(\theta)} \cos(\theta) d\theta \] \[ \int \sin^3(\theta) \sqrt{\cos^2(\theta)} \cos(\theta) d\theta \] \[ \int \sin^3(\theta) \cos^2(\theta) d\theta \] \[ \int \sin^3(\theta) (1 - \sin^2(\theta)) d\theta \] \[ \int (\sin^3(\theta) - \sin^5(\theta)) d\theta \] This integral can be solved using standard trigonometric integral formulas. After integrating, don't forget to revert back to the original variable \( x \) using the original substitution.
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