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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 1

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 1

Answers (1)

Answer: 

\frac{x}{2}-\frac{1}{8} \sin (4 x+10)+C
Hint:

Use,\sin^{2}x=\left ( \frac{1-\cos 2x}{2} \right )
Given:

Let I=\int \sin ^{2}(2 x+5) d x

Solution:

I=\int \sin ^{2}(2 x+5) d x

On Substituting the above formula,

So the equation becomes,

\Rightarrow \int \frac{1-\cos 2\left ( 2x+5 \right )}{2}dx

We know,

\begin{aligned} &\int \cos \mathrm{axdx}=\frac{1}{\mathrm{a}} \sin \mathrm{ax}+\mathrm{C} \\ &\Rightarrow \frac{1}{2} \int \mathrm{dx}-\frac{1}{2} \int \cos (4 \mathrm{x}+10) \mathrm{dx} \end{aligned}

 

On integrating we get,

 

\Rightarrow \frac{x}{2}-\frac{1}{8} \sin (4 x+10)+C

So the answer is

\frac{x}{2}-\frac{1}{8} \sin (4 x+10)+C

Posted by

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