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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 4

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 4

Answers (1)

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Answer:  

\frac{x}{2}-\frac{1}{4 b} \sin (2 b x)+C

Hint:

Use \sin ^{2} \mathrm{x}=\left(\frac{1-\cos 2 \mathrm{x}}{2}\right) and

Given:

Let   \mathrm{I}=\int \sin ^{2} \mathrm{bxdx}
Solution:

\mathrm{I}=\int \sin ^{2} \mathrm{bxdx}

Substituting the above formula, we get

I=\int \frac{1-\cos(2bx)}{2}dx

We know,

\begin{aligned} &\int \cos a x d x=\frac{1}{a} \sin a x+C \\ &\Rightarrow \frac{1}{2} \int d x-\frac{1}{2} \int \cos (2 bx) d x \end{aligned}

 

On Integration,

 

\Rightarrow \frac{x}{2}-\frac{1}{4 b} \sin (2 b x)+c

So the answer is

\frac{x}{2}-\frac{1}{4 b} \sin (2 b x)+C

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