How do you find the integral of $e^{2 x} cos 3 x d x$ $e^(2x) cos3x dx$ ?

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How do you find the integral of $e^{2 x} cos 3 x d x$ $e^(2x) cos3x dx$ ?

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Answer 1 · 2018-03-26T10:26:17

$\int e^{2 x} cos 3 x d x = \frac{e^{2 x}}{\sqrt{13}} cos (2 x - {tan}^{- 1} (\frac{3}{2})) + c$ $inte^(2x)cos3xdx=e^(2x)/sqrt13 cos(2x-tan^-1(3/2)) +c$

Explanation:

We know that,

$\int e^{a x} cos b x d x = \frac{e^{a x}}{\sqrt{a^{2} + b^{2}}} cos (b x - θ) + c$ $color(red)(inte^(ax)cosbx dx=e^(ax)/(sqrt(a^2+b^2))cos(bx-theta)+c$

where, $cos θ = \frac{a}{\sqrt{a^{2} + b^{2}}} and sin θ = \frac{b}{\sqrt{a^{2} + b^{2}}}$ $color(red)(costheta=a/(sqrt(a^2+b^2)) and sintheta=b/(sqrt(a^2+b^2))$

Substituting , $a = 2, and b = 3$ $a=2,and b=3$ , we get

$cos θ = \frac{2}{\sqrt{2^{2} + 3^{2}}} = \frac{2}{\sqrt{13}} and sin θ = \frac{3}{\sqrt{2^{2} + 3^{2}}} = \frac{3}{\sqrt{13}}$ $costheta=2/sqrt(2^2+3^2)=2/sqrt13 and sintheta=3/sqrt(2^2+3^2) =3/sqrt13$

$\Rightarrow tan θ = \frac{sin θ}{cos θ} = \frac{\frac{3}{\sqrt{13}}}{\frac{2}{\sqrt{13}}} = \frac{3}{2} \Rightarrow θ = {tan}^{- 1} (\frac{3}{2})$ $=>tantheta=sintheta/costheta = (3/sqrt13)/(2/(sqrt13))=3/2=>theta=tan^-1(3/2)$

So,

$\int e^{2 x} cos 3 x d x = \frac{e^{2 x}}{\sqrt{13}} cos (2 x - {tan}^{- 1} (\frac{3}{2})) + c$ $inte^(2x)cos3xdx=e^(2x)/sqrt13 cos(2x-tan^-1(3/2)) +c$

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How do you find the integral of $e^{2 x} cos 3 x d x$ $e^(2x) cos3x dx$ ?

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