What is the integral of $\frac{{cos}^{2} x}{sin x}$ $(cos^2x) / sinx$ ?

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Answer 1 · 2016-03-29T15:01:20

$\int \frac{{cos}^{2}}{sin x} d x = l n (tan (\frac{x}{2})) + cos x + C$ $int cos^2/sin x d x=l n(tan(x/2))+cos x+C$

$\int \frac{{cos}^{2}}{sin x} d x = ?$ $int cos^2/sin x d x= ?$

${cos}^{2} x = 1 - {sin}^{2} x$ $cos^2 x=1-sin^2x$

$\int \frac{1 - {sin}^{2} x}{sin x} d x = \int (\frac{1}{sin x} - sin x) d x$ $int(1-sin^2 x)/(sin x) d x=int (1/sin x-sin x)d x$

$\int \frac{{cos}^{2}}{sin x} d x = \int \frac{d x}{sin x} - \int sin x d x$ $int cos^2/sin x d x=int (d x)/sin x- int sin x d x$

$\int \frac{{cos}^{2}}{sin x} d x = l n (tan (\frac{x}{2})) + cos x + C$ $int cos^2/sin x d x=l n(tan(x/2))+cos x+C$

What is the integral of cos2xsinx(cos^2x) / sinx?