Get Answers to all your Questions

header-bg qa

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 1

Answers (1)


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

Hint: \int I \: IIdx=I\int IIdx -\int \left ( \frac{d}{dx}I\int II dx \right )dx

ILATE rule

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

Solution: I=e^{ax}\frac{\sin bx}{b}-a\int e^{ax}\frac{\sin bx}{b}dx

\begin{aligned} &I=\frac{1}{b} e^{a x} \sin b x-\frac{a}{b} \int e^{a x} \sin b x d x \\ &I=\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] \\ &I=\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 \\ &I=\frac{e^{a x}}{b^{2}}[b \sin b x+a \cos b x]-\frac{a^{2}}{b^{2}}I+c \\ &I=\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