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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 5

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 5

Answers (1)

Answer:  

\frac{1}{2}[\mathrm{x}-\sin \mathrm{x}]+\mathrm{C}

Hint:

Use \left[2 \sin ^{2} \frac{x}{2}=1-\cos x\right]

Given:

Let \mathrm{I}=\int \sin ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \\
Solution:

\begin{aligned} &\mathrm{I}=\int \sin ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \\ &=\frac{1}{2} \int 2 \sin ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \end{aligned}

\begin{aligned} &=\frac{1}{2} \int(1-\cos x) d x \\ &=\frac{1}{2}\left[\int d x-\int \cos x d x\right] \\ &=\frac{1}{2}[x-\sin x]+C \end{aligned}

So the answer is =\frac{1}{2}[\mathrm{x}-\sin \mathrm{x}]+\mathrm{C}

 

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