If ∫e2x(cos⁡x+7sin⁡x)dx=e2xg(x)+C where c is a constant of integration, then g(0)+gπ2 is equal to

# If $\int {e}^{2x}\left(\mathrm{cos}x+7\mathrm{sin}x\right)dx={e}^{2x}g\left(x\right)+C$ where $c$ is a constant of integration, then $g\left(0\right)+g\left(\frac{\pi }{2}\right)$ is equal to

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

### Solution:

We know that

$\begin{array}{l}\int {e}^{ax}\mathrm{sin}bxdx=\frac{{e}^{ax}}{{a}^{2}+{b}^{2}}\left(a\mathrm{sin}bx-b\mathrm{cos}bx\right)+C\\ \int {e}^{ax}\mathrm{cos}bxdx=\frac{{e}^{ax}}{{a}^{2}+{b}^{2}}\left(a\mathrm{cos}bx+b\mathrm{sin}bx\right)+C\\ \int {e}^{2x}\left(\mathrm{cos}x+7\mathrm{sin}x\right)dx\\ =\int {e}^{2x}\mathrm{cos}xdx+7\int {e}^{2x}\mathrm{sin}xdx\\ =\frac{{e}^{2x}}{5}\left(2\mathrm{cos}x+\mathrm{sin}x\right)+\frac{7}{5}{e}^{2x}\left(2\mathrm{sin}x-\mathrm{cos}x\right)+C\\ =\frac{{e}^{2x}}{5}\left(-5\mathrm{cos}x+15\mathrm{sin}x\right)+C\\ \begin{array}{l}\therefore g\left(x\right)=\frac{1}{5}\left(-5\mathrm{cos}x+15\mathrm{sin}x\right)=\left(-\mathrm{cos}x+3\mathrm{sin}x\right)\\ ⇒ g\left(0\right)+g\left(\frac{\pi }{2}\right)=-1+3=2\end{array}\end{array}$

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)