How do you integrate $\int \frac{cos x}{{(sin x)}^{\frac{1}{2}}} d x$ $int cos x / ((sin x)^(1/2))dx$ ?

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How do you integrate $\int \frac{cos x}{{(sin x)}^{\frac{1}{2}}} d x$ $int cos x / ((sin x)^(1/2))dx$ ?

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Answer 1 · 2016-01-02T23:34:55

$\int \frac{cos x}{{(sin x)}^{\frac{1}{2}}} d x = 2 {(sin x)}^{\frac{1}{2}} + C$ $int cosx/(sinx)^(1/2) dx=2(sinx)^(1/2)+C$

Explanation:

Let's try to guess what was derived.

We could notice, that:
$\frac{d}{d x} sin x = cos x$ $d/dx sinx=cosx$ and $\frac{d}{d x} ({f (x)}^{\frac{1}{2}}) = \frac{\frac{d}{d x} f (x)}{2 {f (x)}^{\frac{1}{2}}}$ $d/dx (f(x)^(1/2))=(d/dx f(x))/(2f(x)^(1/2))$ (chain rule)

So we try to substitute $u = sin x$ $u=sinx$ :
$d u = cos x d x$ $du=cosx dx$

$\int \frac{cos x}{{(sin x)}^{\frac{1}{2}}} d x = \int \frac{d u}{u^{\frac{1}{2}}} = 2 \int \frac{d u}{2 u^{\frac{1}{2}}} = 2 u^{\frac{1}{2}} + C$ $int cosx/(sinx)^(1/2) dx=int (du)/u^(1/2)=2int (du)/(2u^(1/2))=2u^(1/2)+C$

We need to substitute $u = sin x$ $u=sinx$ back:

$2 u^{\frac{1}{2}} + C = 2 {(sin x)}^{\frac{1}{2}} + C$ $2u^(1/2)+C=2(sinx)^(1/2)+C$

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How do you integrate $\int \frac{cos x}{{(sin x)}^{\frac{1}{2}}} d x$ $int cos x / ((sin x)^(1/2))dx$ ?

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Explanation:

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