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

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Answer 1 · 2016-05-22T11:03:44

$= cos (x) + ln ∣ ∣ tan (\frac{x}{2}) ∣ ∣$ $=cos (x)+\ln |tan (frac{x}{2})|$

$\int \frac{{cos}^{2} (x)}{sin (x)} d x$ $\int \frac{\cos ^2(x)}{\sin (x)}dx$

We know,
${cos}^{2} (x) = 1 - {sin}^{2} (x)$ $\cos ^2(x)=1-\sin ^2(x)$

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

Applying sum rule,
$\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x$ $\int f(x)\pm g(x)dx=\int f(x)dx\pm \int g(x)dx$

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

Also,we know,
$\int \frac{1}{sin (x)} d x = ln ∣ ∣ tan (\frac{x}{2}) ∣ ∣$ $\int \frac{1}{\sin (x)}dx=\ln |\tan (\frac{x}{2})\|$

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

now,
$= ln ∣ ∣ tan (\frac{x}{2}) ∣ ∣ - (- cos (x))$ $=\ln |\tan (\frac{x}{2})|-(-\cos (x))$

simplifying it,
$= cos (x) + ln ∣ ∣ tan (\frac{x}{2}) ∣ ∣$ $=cos (x)+\ln |tan (frac{x}{2})|$

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