Question

Solve $\int {{e^{ax}}\cos (bx + c)dx}$.

Answer 1

Here we assume the whole integral $\int {{e^{ax}}\cos (bx + c)dx}$ as a variable and the solve by integration by parts where using the sequence of ILATE ( Inverse trigonometric, Logarithms , algebraic, trigonometric , exponential) for choosing the value of $u,v$ and then substitute in the integration formula of by parts integration.

• If we have $\int {uvdx}$ then the value of integral is given by $\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)dx} } }$ where $u' = \dfrac{{du}}{{dx}}$ is differentiation of $u$ with respect to $x$.

Complete step-by-step answer:
Let us assume the value of whole integral as $I$
i.e. $I = \int {{e^{ax}}\cos (bx + c)dx}$ $...(i)$
Now, using the ILATE sequence we can tell that the term comes first in the sequence is$u$and which comes later is $v$ i.e. in this case we have the terms exponential and trigonometric , first comes the trigonometric term then comes the exponential term so $u$ is $\cos (bx + c)$ and $v$is ${e^{ax}}$
$u = \cos (bx + c),v = {e^{ax}}$
Now we calculate the differentiation of the term $u$

$$u' = \dfrac{{du}}{{dx}} \\\ u' = \dfrac{{d(\cos (bx + c))}}{{dx}} \\\ u' = - \sin (bx + c) \times \dfrac{{d(bx + c)}}{{dx}} \\\ u' = - b\sin (bx + c) \\\$$

Now substitute the values in equation $\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)dx} } }$
$I = \cos (bx + c)\int {{e^{ax}}dx} - \int {( - b\sin (bx + c))\left( {\int {{e^{ax}}dx} } \right)dx}$ $...(ii)$
Now we know $\int {{e^{mx}} = \dfrac{{{e^{mx}}}}{m}}$ where $m$ is constant.
Therefore substitute the value of $\int {{e^{ax}} = \dfrac{{{e^{ax}}}}{a}}$ in equation $(ii)$
$I = \cos (bx + c) \times \dfrac{{{e^{ax}}}}{a} + b\int {(\sin (bx + c) \times \dfrac{{{e^{ax}}}}{a}dx}$
$I = \dfrac{{{e^{ax}}\cos (bx + c)}}{a} + b\int {\dfrac{{{e^{ax}}(\sin (bx + c)}}{a}dx}$
Now we can take out constant term in the denominator of integration out of the integration
$I = \dfrac{{{e^{ax}}\cos (bx + c)}}{a} + \dfrac{b}{a}\int {{e^{ax}}(\sin (bx + c)dx}$ $...(iii)$
Now we can see there is again an integration left on RHS of the equation, therefore we solve it separately and substitute it back in the equation $(iii)$
To solve $\int {{e^{ax}}(\sin (bx + c)dx}$
We use using the ILATE sequence we can tell that the term comes first in the sequence is$u$and which comes later is $v$ i.e. in this case we have the terms exponential and trigonometric , first comes the trigonometric term then comes the exponential term so $u$ is $\sin (bx + c)$ and $v$is ${e^{ax}}$
$u = \sin (bx + c),v = {e^{ax}}$
Now we calculate the differentiation of the term $u$

$$u' = \dfrac{{du}}{{dx}} \\\ u' = \dfrac{{d(\sin (bx + c))}}{{dx}} \\\ u' = \cos (bx + c) \times \dfrac{{d(bx + c)}}{{dx}} \\\ u' = b\cos (bx + c) \\\$$

Now substitute the values in equation $\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)dx} } }$
$\int {{e^{ax}}(\sin (bx + c)dx} = \sin (bx + c)\int {{e^{ax}}dx} - \int {(b\cos (bx + c))\left( {\int {{e^{ax}}dx} } \right)dx}$ $...(iv)$
Now we know $\int {{e^{mx}} = \dfrac{{{e^{mx}}}}{m}}$ where $m$ is constant.
Therefore substitute the value of $\int {{e^{ax}} = \dfrac{{{e^{ax}}}}{a}}$ in equation $(iv)$

$$\int {{e^{ax}}(\sin (bx + c)dx} = \sin (bx + c) \times \dfrac{{{e^{ax}}}}{a} - \int {b\cos (bx + c) \times \dfrac{{{e^{ax}}}}{a}dx} \\\ \int {{e^{ax}}(\sin (bx + c)dx} = \dfrac{{{e^{ax}}\sin (bx + c)}}{a} - \int {\dfrac{{b{e^{ax}}\cos (bx + c)}}{a}dx} \\\$$

Now we can take out constant term in the denominator of integration out of the integration
$\int {{e^{ax}}(\sin (bx + c)dx} = \dfrac{{{e^{ax}}\sin (bx + c)}}{a} - \dfrac{b}{a}\int {{e^{ax}}\cos (bx + c)dx}$ $...(v)$
From equation $(i)$ we know $I = \int {{e^{ax}}\cos (bx + c)dx}$
Therefore substitute value of $\int {{e^{ax}}\cos (bx + c)dx} = I$ in equation $(v)$
$\int {{e^{ax}}(\sin (bx + c)dx} = \dfrac{{{e^{ax}}\sin (bx + c)}}{a} - \dfrac{b}{a} \times I$
Now we substitute back the value of $\int {{e^{ax}}(\sin (bx + c)dx}$ in equation $(iii)$

I = \dfrac{{{e^{ax}}\cos (bx + c)}}{a} + \dfrac{b}{a}\left[ {\dfrac{{{e^{ax}}\sin (bx + c)}}{a} - \dfrac{b}{a} \times I} \right] \\\ I = \dfrac{{{e^{ax}}\cos (bx + c)}}{a} + \dfrac{b}{a} \times \dfrac{{{e^{ax}}\sin (bx + c)}}{a} - \dfrac{b}{a} \times \dfrac{{bI}}{a} \\\ I = \dfrac{{{e^{ax}}\cos (bx + c)}}{a} + \dfrac{{b{e^{ax}}\sin (bx + c)}}{{{a^2}}} - \dfrac{{{b^2}I}}{{{a^2}}} \\\ $$ { $$\dfrac{b}{a}$$ is multiplied to bracket and $$a \times a = {a^2}$$} Now take LCM of the terms in RHS of the equation $$I = \dfrac{{a{e^{ax}}\cos (bx + c) + b{e^{ax}}\sin (bx + c)}}{{{a^2}}} - \dfrac{{{b^2}I}}{{{a^2}}}$$ Shifting the term with $$I$$ to the LHS of the equation. $$I + \dfrac{{{b^2}I}}{{{a^2}}} = \dfrac{{a{e^{ax}}\cos (bx + c) + b{e^{ax}}\sin (bx + c)}}{{{a^2}}}$$ Taking LCM on LHS of the equation $$\dfrac{{({a^2} + {b^2})I}}{{{a^2}}} = \dfrac{{a{e^{ax}}\cos (bx + c) + b{e^{ax}}\sin (bx + c)}}{{{a^2}}}$$ Now we can cancel out the denominator on both sides of the equations because they are equal. $$({a^2} + {b^2})I = a{e^{ax}}\cos (bx + c) + b{e^{ax}}\sin (bx + c)$$ Take $${e^{ax}}$$ common on RHS of the equation $$({a^2} + {b^2})I = {e^{ax}}\\{ a\cos (bx + c) + b\sin (bx + c)\\} $$ Now to find the value of $$I$$ we divide both sides of the equation by $$({a^2} + {b^2})$$

\dfrac{{({a^2} + {b^2})I}}{{{a^2} + {b^2}}} = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}\{ a\cos (bx + c) + b\sin (bx + c)\} \\
I = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}\{ a\cos (bx + c) + b\sin (bx + c)\} \\

Therefore, value of the integral $$\int {{e^{ax}}\cos (bx + c)dx} = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}\\{ a\cos (bx + c) + b\sin (bx + c)\\} $$ **Note:** Students are likely to make mistake in integration by parts when they don’t do it step by step, that is instead of substituting the value of differentiation after it is done separately they differentiate the term after substituting it into the formula which causes confusion sometimes because the second half of the integration has double integration.