Question

If $\int e^{ax}cos\left(bx\right)dx=\frac{e^{ax}}{K}(acosbx+bsinbx)+C$, then the K here would be equal to -

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

We’ll use integration by parts method to calculate $\int e^{ax}cos\left(bx\right)dx$ being cos(bx) as the first function and $e^{ax}$ being the second function.
Let $I=\int e^{ax}cos\left(bx\right)dx$
Or $I=cos\left(bx\right).\frac{e^{ax}}{a}-\int \frac{e^{ax}}{a}(-sin(bx\left)\right).bdx$
Or $I=cos\left(bx\right).\frac{e^{ax}}{a}+\frac{b}{a}\int e^{ax}\left(sin\right(bx\left)\right).dx$
Let $I_1=\int e^{ax}\left(sin\right(bx\left)\right).dx$
$I=cos\left(bx\right).\frac{e^{ax}}{a}+\frac{b}{a}{I}_1$…(1)
Applying integration by parts to integrate $I_1$, sin(bx) being the first function, and $e^{ax}$ being the second function.
$I_1=sin\left(bx\right).\frac{e^{ax}}{a}-\int \frac{e^{ax}}{a}\left(cos\right(bx\left)\right).bdx$
Or $I_1=sin\left(bx\right).\frac{e^{ax}}{a}-\frac{b}{a}\int e^{ax}\left(cos\right(bx\left)\right).dx$
Let's substitute $I_1=sin\left(bx\right).\frac{e^{ax}}{a}-\frac{b}{a}\int e^{ax}\left(cos\right(bx\left)\right).dx$ in 1st equation.
$I=cos\left(bx\right).\frac{e^{ax}}{a}+\frac{b}{a}\left(sin\right(bx).\frac{e^{ax}}{a}-\frac{b}{a}\int e^{ax}\left(cos\right(bx\left)\right).dx)$
Or $I=cos\left(bx\right).\frac{e^{ax}}{a}+\frac{b}{a^2}sin\left(bx\right)e^{ax}.-\frac{b^2}{a^2}\int e^{ax}\left(cos\right(bx\left)\right).dx$
We can see that in above equation we have $\int e^{ax}cos\left(bx\right)dx$ which is nothing but "I" itself.
$I=cos\left(bx\right).\frac{a^{ax}}{a}+\frac{b}{a^2}sin\left(bx\right).e^{ax}-\frac{b^2}{a^2}$ (1)
Or $I(1+\frac{b^2}{a^2})=cos\left(bx\right).\frac{e^{ax}}{a}+\frac{b}{a^2}sin\left(bx\right).e^{ax}$
Or $I=\frac{a^2}{a^2+b^2}\frac{e^{ax}\left(acos\right(bx)+bsin(bx\left)\right)}{a^2}$
$I=\frac{e^{ax}}{a^2+b^2}\left(acos\right(bx)+bsin(bx)+C$
On comparing we can see that ‘K’ is equal to $a^2+b^2$.