#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 4

Answer: $I=\frac{e^{2 x}}{13}[2 \cos (3 x+4)+3 \sin (3 x+4)]+c$

Hint: Using $\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d x\right) d x$

Given: $\int e^{2 x} \cos (3 x+4) d x$

Solution:

$\text { Let } \mathrm{I}=\int e^{2 x} \cos (3 x+4) d x$

As we know that

\begin{aligned} &\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d x\right) d x\\ &\text { Hence }\\ &f(x)=\cos (3 x+4), g(x)=e^{2 x}\\ &f^{\prime}(x)=-\sin (3 x+4) \times 3=-3 \sin (3 x+4) \end{aligned}

\begin{aligned} &\int g(x)=\int e^{2 x} d x=\frac{e^{2 x}}{2} \\ &I=\cos (3 x+4) \frac{e^{2 x}}{2} \int-3 \sin (3 x+4) \times\left(\frac{e^{2 x}}{2}\right) d x \\ &=\frac{1}{2} e^{2 x} \cos (3 x+4)+\frac{3}{2} \int e^{2 x} \cos (3 x+4) d x \end{aligned}

As we know that

\begin{aligned} &\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d x\right) d x\\ &\text { Hence }\\ &f(x)=\sin (3 x+4), g(x)=e^{2 x}\\ &f^{\prime}(x)=\cos (3 x+4) \times 3=3 \cos (3 x+4)\\ &\int g(x)=\int e^{2 x} d x=\frac{e^{2 x}}{2} \end{aligned}

\begin{aligned} &I=\frac{1}{2} e^{2 x} \cos (3 x+4)+\frac{3}{2} \sin (3 x+4) \frac{e^{2 x}}{2}-\int 3 \cos (3 x+4) d x \\ &I=\frac{1}{2} e^{2 x} \cos (3 x+4)+\frac{3}{4} e^{2 x} \sin (3 x+4)-\frac{9}{4} \int e^{2 x} \cos (3 x+4) d x \\ &=\frac{1}{2} e^{2 x} \cos (3 x+4)+\frac{3}{4} e^{2 x} \sin (3 x+4)-\frac{9}{4} I \end{aligned}

\begin{aligned} &I\left(1+\frac{9}{4}\right)=\frac{1}{2} e^{2 x} \cos (3 x+4)+\frac{3}{4} e^{2 x} \sin (3 x+4) \\ &I\left(\frac{13}{4}\right)=\frac{1}{2} e^{2 x} \cos (3 x+4)+\frac{3}{4} e^{2 x} \sin (3 x+4) \\ &I=\frac{2}{13} e^{2 x} \cos (3 x+4)+\frac{3}{13} e^{2 x} \sin (3 x+4) \\ &I=\frac{e^{2 x}}{13}(2 \cos (3 x+4)+3 \sin (3 x+4)) \end{aligned}