Question

Hi! Please help me out on this: :}
${\sf{\int \: (e ^{ - 3x} {cos}^{3} x) \: dx}}$
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Answer 1

Answer:

${\displaystyle{\pink{\tt \int e^{-3x} \cos^3(x)\ dx = \dfrac{e^{-3x}}{24} \Big(\sin(3x)-\cos(3x)\Big) +\dfrac{3e^{-3x}}{40}\Big(\sin(x)-3\cos(x)\Big) + C}}}$

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Step-by-step explanation:

We have to solve the given indefinite integral.

$\displaystyle\tt I = \int e^{-3x} \cos^3(x)\ dx$

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Using the following trig. identity:

$\star\quad\boxed{\sf{\red{4\cos^3(x) = 3\cos(x) + \cos(3x) }}}$

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$\displaystyle\tt\implies I = \int e^{-3x} \left[\dfrac{\cos(3x) + 3\cos(x)}{4}\right]\ dx$

$\displaystyle\tt\implies I =\underbrace{\tt\dfrac14 \int e^{-3x} \cos(3x) \ dx}_{I_1} + \underbrace{\tt\dfrac34\int e^{-3x}\cos(x)\ dx}_{I_2}$

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Now, we will use the following identity:

${\star\quad\boxed{\sf{\red{\int e^{ax}\cos(bx)\ dx = \dfrac{e^{ax}}{a^2+b^2}\Big(a\cos(bx) +b\sin(bx)\Big) + C}}}}$

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So, using the above formula, we can solve I1.

$\displaystyle\tt\implies I_1 =\dfrac14 \int e^{-3x} \cos(3x) \ dx$

$\displaystyle\tt\implies I_1 =\dfrac14 \cdot\dfrac{e^{-3x}}{(-3)^2+(3)^2} \Big(-3\cos(3x) + 3\sin(3x)\Big) + C$

$\displaystyle\tt\implies I_1 =\dfrac14 \cdot\dfrac{e^{-3x}}{9+9} \Big(-3\cos(3x) + 3\sin(3x)\Big) + C$

$\displaystyle\tt\implies I_1 =\dfrac14 \cdot\dfrac{e^{-3x}}{18} \Big(-3\cos(3x) + 3\sin(3x)\Big) + C$

${\displaystyle\tt\implies I_1 =\dfrac{e^{-3x}}{72} \Big(-3\cos(3x) + 3\sin(3x)\Big) + C\bf \qquad...(1.)}$

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Similarly, evaluating I2 using the formula.

$\displaystyle\tt\implies I_2 =\dfrac34 \int e^{-3x} \cos(x) \ dx$

$\displaystyle\tt\implies I_2 =\dfrac34 \cdot\dfrac{e^{-3x}}{(-3)^2+ (1)^2}\Big(-3\cos(x) + \sin(x)\Big) + C$

$\displaystyle\tt\implies I_2 =\dfrac34 \cdot\dfrac{e^{-3x}}{9+ 1}\Big(-3\cos(x) + \sin(x)\Big) + C$

$\displaystyle\tt\implies I_2 =\dfrac34 \cdot\dfrac{e^{-3x}}{10}\Big(-3\cos(x) + \sin(x)\Big) + C$

${\displaystyle\tt\implies I_2 =\dfrac{3e^{-3x}}{40}\Big(-3\cos(x) + \sin(x)\Big) + C\bf\qquad...(2.)}$

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Adding equation (1.) and (2.) will give us the original integral.

$\displaystyle \tt\longrightarrow I = I_1 + I_2$

$\displaystyle \tt\longrightarrow I = \dfrac{e^{-3x}}{72} \Big(-3\cos(3x) + 3\sin(3x)\Big) +\dfrac{3e^{-3x}}{40}\Big(-3\cos(x) + \sin(x)\Big) + C$

$\boxed{\blue{ \tt I = \dfrac{e^{-3x}}{24} \Big(\sin(3x)-\cos(3x)\Big) +\dfrac{3e^{-3x}}{40}\Big(\sin(x)-3\cos(x)\Big) + C}}$

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This is the final answer.

Answer 2

Answer:

Sorry i couldnt answer ur previous question.

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And whats ur real name, sweety-boi?

Anyways, cheerio!

Have a nice day, eh, sweety boy.

Sayonara

P.S: Report this. Hope i havent wasted ya time.