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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.

Answers (1)

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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infoexpert21

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