int_{0}^{frac{pi}{2}} frac{4x sin x + x^2 cos | vedclass

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Question

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{4x \sin x + x^2 \cos x}{2\sqrt{\sin x}} dx$. We can rewrite the integrand as: $\frac{4x \sin x + x^2 \cos

Vedclass · Accepted Answer

$\frac{\pi^2}{4}$

Vedclass · Answer

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{4x \sin x + x^2 \cos x}{2\sqrt{\sin x}} dx$. We can rewrite the integrand as: $\frac{4x \sin x + x^2 \cos x}{2\sqrt{\sin x}} = 2x \sqrt{\sin x} + \frac{x^2 \cos x}{2\sqrt{\sin x}}$. Notice that this is the derivative of the product $f(x) = x^2 \sqrt{\sin x}$. Using the product rule: $\frac{d}{dx}(x^2 \sqrt{\sin x}) = 2x \sqrt{\sin x} + x^2 \cdot \frac{1}{2\sqrt{\sin x}} \cdot \cos x = 2x \sqrt{\sin x} + \frac{x^2 \cos x}{2\sqrt{\sin x}}$. Thus,the integral becomes: $I = \int_{0}^{\frac{\pi}{2}} \frac{d}{dx}(x^2 \sqrt{\sin x}) dx$. By the Fundamental Theorem of Calculus: $I = [x^2 \sqrt{\sin x}]_{0}^{\frac{\pi}{2}} = ((\frac{\pi}{2})^2 \sqrt{\sin(\frac{\pi}{2})}) - (0^2 \sqrt{\sin(0)}) = \frac{\pi^2}{4} \cdot 1 - 0 = \frac{\pi^2}{4}$.

$\int_{0}^{\frac{\pi}{2}} \frac{4x \sin x + x^2 \cos x}{2\sqrt{\sin x}} dx$ is equal to

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