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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 3

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 3

Answers (1)

Answer:  

\frac{24 x+8 \sin 4 x+\sin 8 x}{64}+C

Hint:

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

Consider \cos ^{4} 2 x d x=\left(\cos ^{2} 2 x\right)^{2}

Given:

Let  \mathrm{I}=\int \cos ^{4} 2 \mathrm{xdx}
Solution:

Considering,

\cos ^{4} 2 x d x=\left(\cos ^{2} 2 x\right)^{2}

The above equation becomes,

\begin{aligned} &\Rightarrow\left(\cos ^{2} 2 \mathrm{x}\right)^{2}=\left(\frac{1+\cos 4 \mathrm{x}}{2}\right)^{2} \\ &\Rightarrow\left(\frac{1+\cos 4 \mathrm{x}}{2}\right)^{2}=\left(\frac{1+2 \cos 4 \mathrm{x}+\cos ^{2} 4 \mathrm{x}}{4}\right) \\ &\Rightarrow \cos ^{2} 4 \mathrm{x}=\frac{1+\cos 8 \mathrm{x}}{2} \\ &\Rightarrow\left(\frac{1+2 \cos 4 \mathrm{x}+\cos ^{2} 4 \mathrm{x}}{4}=\frac{1}{4}+\frac{\cos 4 \mathrm{x}}{2}+\frac{1+\cos 8 \mathrm{x}}{8}\right) \end{aligned}

Then the question becomes,

\Rightarrow \frac{1}{4} \int \mathrm{dx}+\frac{1}{2} \int \cos 4 \mathrm{xdx}+\frac{1}{8} \int \mathrm{dx}+\frac{1}{8} \int \cos 8 \mathrm{xdx}

We know,

\begin{aligned} &\int \operatorname{cosax} d x=\frac{1}{a} \sin a x+c \\ &\Rightarrow \frac{x}{4}+\frac{1}{8} \sin 4 x+\frac{x}{8}+\frac{\sin 8 x}{64}+C \\ &\Rightarrow \frac{24 x+8 \sin 4 x+\sin 8 x}{64} \\ &\frac{24 x+8 \sin 4 x+\sin 8 x}{64}+C \end{aligned}

Posted by

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