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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 10

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 10

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Answer: I=\frac{e^{2x}}{4}+\frac{e^{2x}}{8}\left ( \sin 2x+\cos 2x \right )+c

Hint: Using ILATE rule

Given: \int e^{2x}\left ( \frac{1+\cos 2x}{2} \right )dx

Solution: I=\int e^{2x}\left ( \frac{1+\cos 2x}{2} \right )dx

\begin{aligned} &=\frac{1}{2} \int e^{2 x} d x+\frac{1}{2} \int e^{2 x} \cos 2 x d x \\ &I=\frac{e^{2 x}}{4}+\frac{1}{2} I_{1} \quad \rightarrow(1) \\ &I_{1}=\int e^{2 x} \cos 2 x d x \Rightarrow I=e^{2 x} \frac{\sin 2 x}{2}-\int 2 e^{2 x} \frac{\sin 2 x}{2} d x \\ &=e^{2 x} \frac{\sin 2 x}{2}-\left[e^{2 x}\left(\frac{-\cos 2 x}{2}\right)-\int 2 e^{2 x}\left(\frac{-\cos 2 x}{2}\right) d x\right] \\ &=e^{2 x} \frac{\sin 2 x}{2}+e^{2 x}\left(\frac{\cos 2 x}{2}\right)-\int e^{2 x} \cos 2 x d x \\ &I_{1}+I_{1}=\frac{e^{2 x}}{2}[\sin 2 x+\cos 2 x]+c \\ &I_{1}=\frac{e^{2 x}}{4}[\sin 2 x+\cos 2 x]+c \\ &I=\frac{e^{2 x}}{4}+\frac{e^{2 x}}{8}(\sin 2 x+\cos 2 x)+c[\text { from }(1)] \end{aligned}

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