Question

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

• A. $\sqrt{a^2+b^2}$
• B. $(a^2+b^2{)}^{\frac{1}{3}}$
• C. $a^2+b^2$
• D. $(a^2-b^2)$

Answer 1

The correct option is C $a^2+b^2$

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 $I_1$, $\sin \left(bx\right)$ 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.
$I=\cos \left(bx\right).\frac{e^{ax}}{a}+\frac{b}{a\left(\sin \right(bx\left)\right)}\frac{e^{ax}}{a}-\frac{b}{a}\int e^{ax}\left(\cos \right(bx\left)\right).dx)$
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 $I=\frac{a^2}{a^2+b^2}\frac{e^{ax}\left(a\cos \right(bx)+b\sin (bx\left)\right)}{a^2}$
$I=\frac{e^{ax}}{a^2+b^2}\left(a\cos \right(bx)+b\sin (bx)+C$
On comparing we can see that ‘K’ is equal to $a^2+b^2$.