How do you evaluate the integral #int sinx/(cosx + cos^2x) dx#?

Question

Charles Arocha · Accepted Answer

The integral equals #ln|secx + 1|+ C#

Call the integral #I#. Note that the expression #cosx + cos^2x = cosx(1 + cosx)#. For partial fractions to work, expressions need to be factored the most possible.

#I = int sinx/(cosx(1 + cosx))#

To perform a partial fraction decomposition, we want to get rid of the trigonometric functions if possible. We can do this through a u-substitution. Let #u = cosx#. Then #du = -sinx dx# and #dx = (du)/(-sinx)#.

#I = int sinx/(u(1 + u)) * (du)/(-sinx)#

#I = -int 1/(u(1 + u)) du#

We're now going to use partial fraction decomposition to seperate integrals.

#A/u + B/(u + 1) = 1/(u(u + 1))#

#A(u + 1) + Bu = 1#

#Au + A + Bu = 1#

#(A + B)u + A = 1#

Now write a system of equations.

#{(A + B = 0), (A = 1):}#

This means that #A = 1# and #B = -1#.

The integral becomes.

#I = -int 1/u - 1/(u + 1)du#

#I =-int1/u + int1/(u + 1)du#

#I = ln|u + 1| - ln|u| + C#

#I=ln|cosx + 1| - ln|cosx| + C#

This can be simplified using #lna - lnb = ln(a/b)#.

#I = ln|(cosx + 1)/cosx|#

We can rewrite #1/cosx# as #secx#.

#I = ln|secx(cosx + 1)|#

#I = ln|1 + secx|#, since #secx# and #cosx# are reciprocals.

Hopefully this helps!

Emily Ammerman · Answer

# ln|1+secx|+C.#

Let #I=intsinx/(cosx+cos^2x)dx.#

#"Subst. "u=cosx rArr du=-sinxdx," so, I=-int1/(u+u^2)du.#

We can integrate using Partial Factions, but, it is much simpler

without that.

#I=-int1/(u(1+u))du=-int{(u+1)-u}/(u(u+1))du#

#=-int{(u+1)/(u(u+1)}-u/(u(u+1))}du#

#=int1/(u+1)du-int1/udu#

#=ln|u+1|-ln|u|#

#=ln|(u+1)/u|=ln|u/u+1/u|#

Since, #u=cosx#, we have,

#I=ln|1+secx|+C.#

Enjoy Maths.!

Cameron Alexander · Answer

undefined To evaluate the integral $ \int \frac{\sin(x)}{\cos(x) + \cos^2(x)} \, dx $, perform the substitution $ u = \cos(x) $. Then, $ du = -\sin(x) \, dx $. The integral becomes $ \int \frac{-1}{u + u^2} \, du $. Rewrite the integrand as $ \frac{-1}{u(1 + u)} $. Decompose this using partial fractions as $ \frac{A}{u} + \frac{B}{1 + u} $. Solve for $ A $ and $ B $, then integrate each term separately. After integrating, revert back to the original variable $ x $ to obtain the final solution.