How do you Integrate $cosx/(sinx)^2+sinx$ ?

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Answer 1 · 2015-03-06T22:10:19

The required integral is

$I = int[cosx/(sinx)^2 + sinx]dx$

This can be evaluated as two separate integrals,

$I = I_1 + I_2$ , where $I_1 = intcosx/(sinx)^2dx$ and $I_2 = intsinxdx$

The solution of $I_2$ is trivial,

$I_2 = intsinxdx = -cosx + C_2$

For $I_1:$

Let $sinx = t$

Differentiating with respect to t,

$cosxdx/dt = 1$

$=> cosxdx = dt$

This transforms $I_1$ to

$I_1 = int1/t^2dt$

which has the simple solution of

$I_1 = -1/t + C_1$

Replacing the value of $t = sinx$

$I_1 = -1/sinx + C_1$

Combining both solutions,

$I = I_1 + I_2$
$=> I = -1/sinx + C_1 -cosx + C_2$

The constants of integration $C_1$ and $C_2$ can be added without affecting the solution.

Finally, the answer is

$I = -1/sinx -cosx + C$

How do you Integrate cosx/(sinx)^2+sinxcosx/(sinx)^2+sinx?