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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 12 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 12 Maths Textbook Solution.

Answers (1)

Answer:

\frac{\sin 3 x}{3}-\frac{\sin ^{8} 3 x}{9}+c

Given:

\int \cos ^{3} 3 x d x

Hint:

Let the term and derivate it and then integrate it.

Solution:   

\int \cos ^{3} 3 x d x

       =\int \cos ^{2} 3 x \cos 3 x d x

   =\int\left(1-\sin ^{2} 3 x\right) \cos 3 x d x \quad\left[\because \cos ^{2} x+\sin ^{2} x=1\right]

   now,let \mathrm{u}=\sin 3 \mathrm{x}

   Differentiate w.r.t x

    \frac{\mathrm{du}}{\mathrm{dx}}=3 \cos 3 \mathrm{x}

   d u=3 \cos 3 x d x

   Now,

   \frac{1}{3} \int 3\left(1-\sin ^{2} 3 x\right) \cos 3 x d x

 =\frac{1}{3} \int 1-u^{2} d u \quad(p u t u \& d u)

 =\frac{1}{3} \int d u-\frac{1}{3} \int u^{2} d u

 =\frac{1}{3} u-\frac{1}{3} \times \int u^{2} d u

 =\frac{1}{3} \sin 3 x-\frac{\sin ^{8} 3 x}{9}+c

    

 

Posted by

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