How do you Integrate $\frac{cos x}{{(sin x)}^{2}} + sin x$ $cosx/(sinx)^2+sinx$ ?

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How do you Integrate $\frac{cos x}{{(sin x)}^{2}} + sin x$ $cosx/(sinx)^2+sinx$ ?

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Answer 1 · 2015-03-06T22:10:19

The required integral is

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

This can be evaluated as two separate integrals,

$I = I_{1} + I_{2}$ $I = I_1 + I_2$ , where $I_{1} = \int \frac{cos x}{{(sin x)}^{2}} d x$ $I_1 = intcosx/(sinx)^2dx$ and $I_{2} = \int sin x d x$ $I_2 = intsinxdx$

The solution of $I_{2}$ $I_2$ is trivial,

$I_{2} = \int sin x d x = - cos x + C_{2}$ $I_2 = intsinxdx = -cosx + C_2$

For $I_{1} :$ $I_1:$

Let $sin x = t$ $sinx = t$

Differentiating with respect to t,

$cos x \frac{d x}{d t} = 1$ $cosxdx/dt = 1$

$\Rightarrow cos x d x = d t$ $=> cosxdx = dt$

This transforms $I_{1}$ $I_1$ to

$I_{1} = \int \frac{1}{t^{2}} d t$ $I_1 = int1/t^2dt$

which has the simple solution of

$I_{1} = - \frac{1}{t} + C_{1}$ $I_1 = -1/t + C_1$

Replacing the value of $t = sin x$ $t = sinx$

$I_{1} = - \frac{1}{sin x} + C_{1}$ $I_1 = -1/sinx + C_1$

Combining both solutions,

$I = I_{1} + I_{2}$ $I = I_1 + I_2$
$\Rightarrow I = - \frac{1}{sin x} + C_{1} - cos x + C_{2}$ $=> I = -1/sinx + C_1 -cosx + C_2$

The constants of integration $C_{1}$ $C_1$ and $C_{2}$ $C_2$ can be added without affecting the solution.

Finally, the answer is

$I = - \frac{1}{sin x} - cos x + C$ $I = -1/sinx -cosx + C$

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How do you Integrate $\frac{cos x}{{(sin x)}^{2}} + sin x$ $cosx/(sinx)^2+sinx$ ?

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