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

1 Answer
Mar 6, 2015

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#