Integrate:&nbsp;∫3sin(x)+2cos(x)2sin(x)+3cos(x)dx

Question

Integrate:&amp;nbsp;∫3sin(x)+2cos(x)2sin(x)+3cos(x)dx

Jillian Edgerton · Accepted Answer

We can split this into two integrals:  3∫sin⁡(x)2sin⁡(x)+3cos⁡(x)dx+2∫cos⁡(x)2sin⁡(x)+3cos⁡(x)dx  Focusing on the second integral we find:  ∫cos⁡(x)2sin⁡(x)+3cos⁡(x)dx=∫12tan⁡(x)+3dx  Make the substitution u=2tan⁡(x)+3 which makes  du=2sec2(x)dx=2(tan2(x)+1)dx=2(u2+1)dx  Thus we have  ∫cos⁡(x)2sin⁡(x)+3cos⁡(x)dx=12∫1u(u2+1)du=12∫(1u−uu2+1)du  This integral can be computed by another substitution:  ∫cos⁡(x)2sin⁡(x)+3cos⁡(x)dx=12ln|2tan⁡(x)+3|−12ln|(2tan⁡(x)+3)2+1|+C  That completes the second integral. The first can be handled in a similar manner.

veiga34 · Accepted Answer

It is a pity you did not have a minus-sign in the numerator, since

D \ln (3 \cos x + 2 \sin x) = \frac{2 \cos x - 3 \sin x}{3 \cos x + 2 \sin x}

but let us see how we can use this fact anyways.
Let us aim at writing

\frac{2 \cos x + 3 \sin x}{3 \cos x + 2 \sin x} = c_{1} \frac{2 \cos x - 3 \sin x}{3 \cos x + 2 \sin x} + c_{2} \frac{3 \cos x + 2 \sin x}{3 \cos x + 2 \sin x}

since both those terms are easy to integrate. This leads us to the linear equations

2 = 2 c_{1} + 3 c_{2}

and

3 = - 3 c_{1} + 2 c_{2}

. The solution to this system is

c_{1} = - \frac{5}{13}

and

c_{2} = \frac{12}{13}

. Thus

\int \frac{2 \cos x + 3 \sin x}{3 \cos x + 2 \sin x} d x = - \frac{5}{13} \int \frac{2 \cos x - 3 \sin x}{3 \cos x + 2 \sin x} d x + \frac{12}{13} \int \frac{3 \cos x + 2 \sin x}{3 \cos x + 2 \sin x} d x

.
I guess you can take it from here?

karton · Accepted Answer

Let there be a right triangle with angle A(not with measurement 90 deg) whose adjacent has measurement 3 and whose opposite has measerement 2. Therefore the hypotenuse of the triangle has measurement $\sqrt{13}$
So this means

$\sin (A) = \frac{2}{\sqrt{13}} and \cos (A) = \frac{3}{\sqrt{13}} . 3 \sin (x) + 2 \cos (X) = \sqrt{13} (\frac{3}{\sqrt{13}} \sin (x) + \frac{2}{\sqrt{13}} \cos (x)) = \sqrt{13} (\cos (A) \sin (x) + \sin (A) \cos (x)) = \sqrt{13} \cos (x - A) \int \frac{3 \sin (x) + 2 \cos (x)}{2 \sin (x) + 3 \cos (x)} d x = \int \frac{\sqrt{13} \sin (x + A)}{\sqrt{13} \cos (x - A)} d x = \int \frac{\sin (x - A + 2 A)}{\cos (x - A)} d x = \int \frac{\sin (x - A) \cos (2 A) + \sin (2 A) \cos (x - a)}{\cos (x - A)} = \cos (2 A) \int \frac{\sin (x - A)}{\cos (x - A)} d x + \sin (2 A) \int \frac{\cos (x - A)}{\cos (x - A)} d x = (\cos^{2} (A) - \sin^{2} (A)) \int \frac{\sin (x - A)}{\cos (x - A)} d x + 2 \sin (A) \cos (A) \int 1 d x = ((\frac{3}{\sqrt{13}})^{2} - (\frac{2}{\sqrt{13}})^{2}) \int \frac{- d u}{u} + 2 \frac{2}{\sqrt{13}} \frac{3}{\sqrt{13}} x + C (note: where u = \cos (x - A) and so d u = - \sin (x - A) d x) = (\frac{9}{13} - \frac{4}{13}) (- \ln | u |) + \frac{12}{13} x + C = - \frac{5}{13} \ln | \cos (x - A) | + \frac{12}{13} x + C = - \frac{5}{13} \ln | \cos (x) \cos (A) + \sin (x) \sin (A) | + \frac{12}{13} x + C = - \frac{5}{13} \ln | \frac{1}{\sqrt{13}} | - \frac{5}{13} \ln | 3 \cos (x) + 2 \sin (x) | + \frac{12}{13} x + C = - \frac{5}{13} \ln | 3 \cos (x) + 2 \sin (x) | + \frac{12}{13} x - \frac{5}{13} \ln | \frac{1}{\sqrt{13}} | + C = - \frac{5}{13} \ln | 3 \cos (x) + 2 \sin (x) | + \frac{12}{13} x + K$

Integrate: int(3sin(x)+2cos(x))(2sin(x)+3cos(x))dx

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