#### Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.18 Question 8 Maths Textbook Solution.

Answer:$\frac{1}{4} \log \left|\sin ^{2} 4 x+\sqrt{9+\sin ^{4} 4 x}\right|+c$

Hint Let $\sin ^{2} 4 x=t$

Given: $\int \frac{\sin 8 x}{\sqrt{9+\sin ^{4} 4 x}} d x$

Put

$\sin^{2}4x=t\Rightarrow 2 \times4\sin 4x \cdot \cos 4x dx=dt$

$\sin 8x dx=\frac{1}{4}dt$

$\\ \\ \Rightarrow \hspace{1cm}I=\frac{1}{4}\int \frac{1}{\sqrt{9+t^{2}}}dt=\log\left | t+\sqrt{9+t^{2}}+ \right |+c\\ \\ \Rightarrow \hspace{1cm}I=\frac{1}{4}\log \left | \sin^{2}4x+\sqrt{9+\sin^{4}4x} \right |+c$

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Answer: $\frac{1}{4} \log \left|\sin ^{2} 4 x+\sqrt{9+\sin ^{4} 4 x}\right|+c$

Hint Let $\sin ^{2} 4 x=t$

Given: $\int \frac{\sin 8 x}{\sqrt{9+\sin ^{4} 4 x}} d x$

Explanation:

$\int \frac{\sin 8 x}{\sqrt{9+\sin ^{4} 4 x}} d x$                ............(1)

Let $\sin ^{2} 4 x=t$

$4 \times 2 \sin 4 x \cos 4 x d x=d t$

$4 \sin 8 x d x=d t$                                                (Differentiate w.r.t to t)

Put in (1) we have

$\frac{1}{4} \int \frac{d t}{\sqrt{9+t^{2}}}=\frac{1}{4} \log \left|t+\sqrt{9+t^{2}}\right|+c \quad\left[\because \int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]$

$=\frac{1}{4} \log \left|\sin ^{2} 4 x+\sqrt{9+\sin ^{4} 4 x}\right|+c$