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# Find: integral of sin³(x)cos(x/2)

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Find: integral of sin³(x)cos(x/2)

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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