What is the integral of $\int \frac{sec x}{{tan}^{2} x} d x$ $int secx/tan^2x dx$ ?

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What is the integral of $\int \frac{sec x}{{tan}^{2} x} d x$ $int secx/tan^2x dx$ ?

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Answer 1 · 2018-06-14T16:16:57

$- csc x + C$ $-cscx+C$

Explanation:

Let's rewrite the integrand:

$\int \frac{sec x}{{tan}^{2} x} d x = \int \frac{\frac{1}{cos x}}{\frac{{sin}^{2} x}{{cos}^{2} x}} d x = \int \frac{cos x}{{sin}^{2} x}$ $intsecx/tan^2xdx=int(1/cosx)/(sin^2x/cos^2x)dx=intcosx/sin^2x$

Here, recognize that $\int \frac{cos x}{{sin}^{2} x} d x = \int \frac{cos x}{sin x} \frac{1}{sin x} d x = \int cot x csc x d x = - csc x + C$ $intcosx/sin^2xdx=intcosx/sinx1/sinxdx=intcotxcscxdx=-cscx+C$ .

Another way we can solve $\int \frac{cos x}{{sin}^{2} x} d x$ $intcosx/sin^2xdx$ is by letting $u = sin x$ $u=sinx$ so $d u = cos x d x$ $du=cosxdx$ . Then,

$\int \frac{cos x}{{sin}^{2} x} d x = \int u^{- 2} d u = - u^{- 1} + C = - \frac{1}{sin x} + C = - csc x + C$ $intcosx/sin^2xdx=intu^-2du=-u^-1+C=-1/sinx+C=-cscx+C$ .

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What is the integral of $\int \frac{sec x}{{tan}^{2} x} d x$ $int secx/tan^2x dx$ ?

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Explanation:

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What is the integral of $\int \frac{sec x}{{tan}^{2} x} d x$ $int secx/tan^2x dx$ ?