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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 11 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise Revision Exercise Question 11 Maths Textbook Solution.

Answers (1)

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Answer:

\frac{3 x}{8}+\frac{\sin 8 x}{64}-\frac{\sin 4 x}{8}+c

Given:

\int \sin ^{4} 2 x d x

Hint:

You must know about trigonometric identities.

Solution:   

\int\left[\sin ^{2}(2 x)\right]^{2} d x

     =\int \frac{(1-\cos 4 x)^{2}}{4} d x \quad \because\left[\cos 2 x=1-2 \sin ^{2} x\right]

     =\int \frac{1-2 \cos 4 x+\cos ^{2} 4 x}{4} d x

     =\int \frac{1-2 \cos 4 x}{4} d x+\int \frac{1+\cos 8 x}{8} d x:\left[1+\cos 2 x=2 \cos ^{2} x\right]

     =\frac{1}{4} \int d x-\frac{1}{2} \int \cos 4 x d x+\frac{1}{8} \int d x+\frac{1}{8} \int \cos 8 x d x

     =\frac{3}{8} x-\frac{\sin 4 x}{8}+\frac{\sin 8 x}{64}+c

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