$\int \frac{\cos ^{3} x}{\sin x} d x=\ln |\sin x|+\frac{\sin ^{2} x}{2}+C$ .

Question

Quizplus · Accepted Answer

The correct integral of $\frac{\cos^3 x}{\sin x} dx$ involves using trigonometric identities and substitution to simplify the integral, which typically results in a different expression. The given integral simplifies to a combination of a logarithmic function and a trigonometric function, but not in the form provided. The correct antiderivative of $\frac{\cos^3 x}{\sin x} dx$ would involve breaking down $\cos^3 x$ into $\cos x \cdot \cos^2 x$ and then using the identity $\cos^2 x = 1 - \sin^2 x$, leading to an integral that can be solved using u-substitution with $u = \sin x$, not directly resulting in the expression given.

∫cos⁡3xsin⁡xdx=ln⁡∣sin⁡x∣+sin⁡2x2+C\int \frac{\cos ^{3} x}{\sin x} d x=\ln |\sin x|+\frac{\sin ^{2} x}{2}+C∫sinxcos3x​dx=ln∣sinx∣+2sin2x​+C

$\int \frac{\cos ^{3} x}{\sin x} d x=\ln |\sin x|+\frac{\sin ^{2} x}{2}+C$