Question

Evaluate $\int 3^x\cos 4xdx=$

• A. $\frac{3^x}{(\log 3{)}^2+16}\left[\right(\log 3)\cos 4x+4\sin 4x]+c$
• B. $\frac{3^x}{9+(\log 4{)}^2}[\cos 4x+(\log 3)\cdot \sin 4x]+c$
• C. $\frac{3^x}{(\log 3{)}^2+(\log 4{)}^2}\left[\right(\log 3)\sin 4x+(\log 4\left)\cos 3x\right]+c$
• D. $\frac{3^x}{(\log 3{)}^2+16}\left[\right(\log 3)\sin 4x+(\log 4\left)\cos 3x\right]+c$

Answer 1

The correct option is A $\frac{3^x}{(\log 3{)}^2+16}\left[\right(\log 3)\cos 4x+4\sin 4x]+c$

$L=\int 3^x\cos 4xdx.$ ...... $\left(i\right)$

$=\cos 4x\int 3^{x}dx+\int 4\sin 4x\frac{3^x}{\left(\log 3\right)}dx.$

$=\frac{3^x\cos 4x}{\log \left(3\right)}+\frac{4}{\log 3}\int 3^x\sin 4xdx$ ...... $\left(ii\right)$

Now, $\int 3^x\sin 4xdx=\frac{\sin 4x{3}^x}{\log 3}-\int 4\cos 4x\cdot \frac{3^x}{\log 3}$

$=\frac{3^x\sin 4x}{\log 3}-\frac{4}{\log 3}\int 3^x\cos 4xdx$

$=\frac{3^x\sin 4x}{\log 3}-\frac{4}{\log 3}L$ ..... From $\left(i\right)$

$\therefore L=\frac{3^x\cos 4x}{\log 3}+\frac{4}{\log 3}[\frac{3^x\sin 4x}{\log 3}-\frac{4}{\log 3}\times L]$ ..... From $\left(ii\right)$

$L\left(\frac{{\left(\log 3\right)}^2+4^2}{{\left(\log 3\right)}^2}\right)=\frac{3^x}{{\left(\log 3\right)}^2}[\cos 4x\left(\log 3\right)+4\sin 4x]$

$L=\frac{3^x}{{\left(\log 3\right)}^2+4^2}[\left(\log 3\right)\cos 4x+4\sin 4x]+c$