What is $\int cot x cos x d x$ $int cotxcosx dx$ ?

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Answer 1 · 2018-04-13T12:54:08

The answer is $= cos x - ln (| csc x + cot x |) + C$ $=cosx-ln(|cscx+cotx|)+C$

$Reminder$ $"Reminder"$

$cot x = \frac{cos x}{sin x}$ $cotx=cosx/sinx$

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

$\int csc x d x = - ln (csc x + cot x)$ $intcscxdx=-ln(cscx+cotx)$

Therefore,

The integral is

$I = \int cot x cos x d x = \int \frac{cos x}{sin x} \cdot cos x d x$ $I=intcotxcosxdx=intcosx/sinx*cosxdx$

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

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

$= \int \frac{d x}{sin x} - \int sin x d x$ $=intdx/sinx-intsinxdx$

$= \int csc x d x + cos x$ $=intcscxdx+cosx$

$= - ln (| csc x + cot x |) + cos x + C$ $=-ln(|cscx+cotx|)+cosx+C$

What is ∫cotxcosxdxint cotxcosx dx?