Question

Question #0bb11

Answer 1

`arctan(2tan(x/2)+1)+C`

Explanation:

`I=int1/(3+2sin(x)-cos(x))dx`

What we will use here is the tangent half-angle substitution, also sometimes known as the Weierstrass substitution.

This substitution allows us to write functions like `sin(x)` and `cos(x)` using `tan(x/2)`, which may seem pointless, but can actually be very useful.

First, we can derive the identities we will need. Using the sine double-angle formula, we can first write that:

`sin(2alpha)=2sin(alpha)cos(alpha)`

Which can be modified to say that:

`sin(x)=2sin(x/2)cos(x/2)`

Rearranging to make tangent "appear:"

`sin(x)=2(sin(x/2)/cos(x/2))cos^2(x/2)=2tan(x/2)(1/sec^2(x/2))`

Using the Pythagorean identity:

`sin(x)=(2tan(x/2))/(1+tan^2(x/2))" "color(red)((star)`

Using a similar process for cosine, starting with its double-angle formula adapted for sine:

`cos(2alpha)=1-2sin^2(alpha)=>cos(x)=1-2sin^2(x/2)`

Continuing on by making tangent "appear:"

`cos(x)=1-2(sin^2(x/2)/cos^2(x/2))cos^2(x/2)=1-2tan^2(x/2)(1/sec^2(x/2))`

`cos(x)=1-(2tan^2(x/2))/(1+tan^2(x/2))`

Once you find a common denominator and combine the fractions:

`cos(x)=(1-tan^2(x/2))/(1+tan^2(x/2))" "color(blue)((star)`

Substituting identities `color(red)((star)` and `color(blue)((star)` into `I`:

`I=int1/(3+(4tan(x/2))/(1+tan^2(x/2))-(1-tan^2(x/2))/(1+tan^2(x/2)))dx`

Finding a common denominator:

`I=int1/((3+3tan^2(x/2))/(1+tan^2(x))+(4tan(x/2))/(1+tan^2(x/2))-(1-tan^2(x/2))/(1+tan^2(x/2)))dx`

`I=int1/((4tan^2(x/2)+4tan(x/2)+2)/(1+tan^2(x/2))dx`

`I=1/2int(1+tan^2(x/2))/(2tan^2(x/2)+2tan(x/2)+1)dx`

Now, let `u=tan(x/2)`. Differentiating this shows that `du=sec^2(x/2)/2dx=(1+tan^2(x/2))/2dx`. This is already contained within the integral!

`I=int(du)/(2u^2+2u+1)=1/2int(du)/(u^2+u+1/2)`

Completing the square gives:

`I=1/2int(du)/((u+1/2)^2+1/4)`

Now, we can treat this like a trig substitution and let `u+1/2=1/2tan(theta)`. This implies that `du=1/2sec^2(theta)d theta`. Substituting in:

`I=1/2int(1/2sec^2(theta)d theta)/((1/2tan(theta))^2+1/4)`

`I=1/4int(sec^2(theta)d theta)/(1/4tan^2(theta)+1/4)`

`I=1/4int(4sec^2(theta)d theta)/(tan^2(theta)+1)`

Since `tan^2(theta)+1=sec^2(theta)`, a fact we've seen about `20` times:

`I=int(sec^2(theta)d theta)/sec^2(theta)=intd theta=theta+C`

From `u+1/2=1/2tan(theta)` we see that `theta=arctan(2u+1)`:

`I=arctan(2u+1)+C`

Since `u=tan(x/2)`:

`I=arctan(2tan(x/2)+1)+C`