Get Answers to all your Questions

header-bg qa

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 32

Answers (1)

Answer: \frac{e^{ax}}{a^{2}+b^{2}}\left [ b\cos bx+a\cos bx \right ]+c

Hints: You must know about the integral rule of trigonometric functions

Given \int e^{ax}\cos bx dx


\int e^{ax}\cos bx dx

Integrating by parts

\begin{aligned} &I=e^{a x} \cdot \frac{\sin b x}{b}-a \int e^{a x} \frac{\sin b x}{b} d x \\ &=\frac{1}{b} e^{a x} \sin b x-\frac{a}{b} \int e^{a x} \sin b x d x \end{aligned}

Again using integration by parts

\begin{aligned} &\frac{1}{b} e^{a x} \sin b x-\frac{a}{b}\left[-e^{a x} \frac{\cos b x}{b}-a \int e^{a x} \frac{\cos b x}{b} d x\right] \\ &\frac{1}{b} e^{a x} \sin b x-\frac{a}{b^{2}} e^{a x} \cos b x-\frac{a^{2}}{b^{2}} \int e^{a x} \cos b x d x \end{aligned}

On computing,

\begin{aligned} &I=\frac{e^{a x}}{b^{2}}[b \sin b x+a \cos b x]-\frac{a^{2}}{b^{2}} I+c \\ &=\frac{e^{a x}}{a^{2}+b^{2}}[b \sin b x+a \cos b x]+c \end{aligned} 

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support