Question #a39e8

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Question #a39e8

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Answer 1 · 2017-06-06T18:17:34

$\int (\frac{3 + 4 sin (x) + 2 cos (x)}{3 + 2 sin (x) + cos (x)}) d x = 2 x - 3 arctan (tan (\frac{1}{2} x) + 1) + K$ $\int ((3+4sin(x)+2cos(x))/(3+2sin(x)+cos(x))) "d"x = 2x - 3arctan(tan(1/2x)+1) + K$ .

Explanation:

$\int (\frac{3 + 4 sin (x) + 2 cos (x)}{3 + 2 sin (x) + cos (x)}) d x$ $\int ((3+4sin(x)+2cos(x))/(3+2sin(x)+cos(x))) "d"x$ .

Observe that the numerator and the denominator are very similar. The numerator is almost exactly double the denominator.

Therefore, add 3 and take away 3 from the numerator in order to lead to nice cancellation.

$\int (\frac{6 + 4 sin (x) + 2 cos (x) - 3}{3 + 2 sin (x) + cos (x)}) d x$ $\int ((6+4sin(x)+2cos(x)-3)/(3+2sin(x)+cos(x))) "d"x$ ,
$\int (\frac{2 (3 + 2 sin (x) + cos (x))}{3 + 2 sin (x) + cos (x)}) - \frac{3}{3 + 2 sin (x) + cos (x)} d x$ $\int ((2(3+2sin(x)+cos(x)))/(3+2sin(x)+cos(x)))-3/(3+2sin(x)+cos(x)) "d"x$ ,
$\int (2 - \frac{3}{3 + 2 sin (x) + cos (x)}) d x$ $\int (2 -3/(3+2sin(x)+cos(x))) "d"x$ ,
$2 x - 3 \int \frac{1}{3 + 2 sin (x) + cos (x)} d x$ $2x - 3 \int 1/(3+2sin(x)+cos(x)) "d"x$ .

Now we just have to solve,

$\int \frac{1}{3 + 2 sin (x) + cos (x)} d x$ $\int 1/(3+2sin(x)+cos(x)) "d"x$ .

Often a useful substitution for tricky trigonometric integrals is a substitution called the tangent half angle substitution or the Weierstrass substitution. It is particularly useful for dealing with rational functions of $sin (x)$ $sin(x)$ and $cos (x)$ $cos(x)$ , as we have here.

The substitution is the following,

$t = tan (\frac{1}{2} x)$ $t = tan(1/2x)$ .

With this substitution,

$\frac{d t}{d x} = \frac{1}{2} \cdot {sec}^{2} (\frac{1}{2} x)$ $("d"t)/("d"x) = 1/2 * sec^2(1/2x)$ .

Using the identity ${sec}^{2} (θ) = 1 + {tan}^{2} (θ)$ $sec^2(theta)=1+tan^2(theta)$ ,

$\frac{d t}{d x} = \frac{1}{2} \cdot (1 + {tan}^{2} (\frac{1}{2} x))$ $("d"t)/("d"x) = 1/2 * (1+tan^2(1/2x))$ ,
$\frac{d t}{d x} = \frac{1}{2} \cdot (1 + t^{2})$ $("d"t)/("d"x) = 1/2 * (1+t^2)$ .

We conclude that,

$d x = \frac{2}{1 + t^{2}} d t$ $"d"x = 2/(1+t^2) "d"t$ .

Consider by the half angle identity for sine,

$sin (x) = 2 sin (\frac{x}{2}) cos (\frac{x}{2})$ $sin(x)=2sin(x/2)cos(x/2)$
$sin (x) = 2 {cos}^{2} (\frac{x}{2}) \cdot (\frac{sin (\frac{x}{2})}{cos (\frac{x}{2})})$ $sin(x)=2cos^2(x/2)*(sin(x/2)/cos(x/2))$
$sin (x) = \frac{2 tan (\frac{x}{2})}{{sec}^{2} (x)}$ $sin(x)=(2tan(x/2))/(sec^2(x))$
$sin (x) = 2 \frac{t}{1 + t^{2}}$ $sin(x)=2t/(1+t^2)$ .

Similarly,

$cos (x) = {cos}^{2} (\frac{x}{2}) - {sin}^{2} (\frac{x}{2})$ $cos(x) = cos^2(x/2)-sin^2(x/2)$
$cos (x) = {cos}^{2} (\frac{x}{2}) ⎛ ⎝ 1 - {(\frac{sin (\frac{x}{2})}{cos (\frac{x}{2})})}^{2} ⎞ ⎠$ $cos(x) = cos^2(x/2) (1-((sin(x/2))/cos(x/2))^2)$
$cos (x) = \frac{1 - {tan}^{2} (\frac{x}{2})}{{sec}^{2} (\frac{x}{2})}$ $cos(x) = (1-tan^2(x/2))/(sec^2(x/2))$
$cos (x) = \frac{1 - t^{2}}{1 + t^{2}}$ $cos(x) = (1-t^2)/(1+t^2)$ .

Substituting, $\int \frac{1}{3 + 2 sin (x) + cos (x)} d x$ $\int 1/(3+2sin(x)+cos(x)) "d"x$ , is transformed to,

$\int \frac{1}{3 + \frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}} \cdot \frac{2}{1 + t^{2}} d t$ $\int 1/(3+(2t)/(1+t^2)+(1-t^2)/(1+t^2)) * 2/(1+t^2) "d"t$
$2 \int \frac{1}{3 (1 + t^{2}) + 2 t + 1 - t^{2}} d t$ $2\int 1/(3(1+t^2)+2t+1-t^2) "d"t$
$2 \int \frac{1}{2 t^{2} + 2 t + 4} d t$ $2\int 1/(2t^2+2t+4)"d"t$
$\int \frac{1}{t^{2} + 2 t + 2} d t$ $\int 1/(t^2+2t+2) "d"t$ .

The quadratic in $t$ $t$ in the denominator is irreducible (cannot be factored without the use of complex numbers), meaning that this is an arctangent integral in $t$ $t$ .

$\int \frac{1}{{(1 + t)}^{2} + 1} d t = arctan (t + 1) + C$ $\int1/((1+t)^2+1) "d"t = arctan(t+1) + C$

Then, reversing the substitution,

$\int \frac{1}{3 + 2 sin (x) + cos (x)} d x = arctan (tan (\frac{1}{2} x) + 1) + C$ $\int 1/(3+2sin(x)+cos(x)) "d"x = arctan(tan(1/2x)+1)+C$ .

We conclude that,

$\int (\frac{3 + 4 sin (x) + 2 cos (x)}{3 + 2 sin (x) + cos (x)}) d x = 2 x - 3 arctan (tan (\frac{1}{2} x) + 1) + K$ $\int ((3+4sin(x)+2cos(x))/(3+2sin(x)+cos(x))) "d"x = 2x - 3arctan(tan(1/2x)+1) + K$ .

Incidentally, you can read more about the tangent half angle substitution here.

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