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

$-\frac{1}{2}$

Given:

$\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x=a \sin 2 x+C$

Hint:

Using identity $\left(a^{2}-b^{2}\right) \& \int \cos x d x$

Explanation:

Let $I=\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x$

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

\begin{aligned} &=\int \frac{\left(\sin ^{4} x-\cos ^{4} x\right)\left(\sin ^{4} x+\cos ^{4} x\right)}{\left(\sin ^{4} x+\cos ^{4} x\right)} d x \\ &=\int\left(\sin ^{4} x-\cos ^{4} x\right) d x \\ &=-\int\left(\cos ^{4} x-\sin ^{4} x\right) d x \\ &=-\int\left(\cos ^{2} x-\sin ^{2} x\right)\left(\cos ^{2} x+\sin ^{2} x\right) d x \end{aligned}                                               $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

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

According to given : $I=a \sin 2 x+C$

$\therefore a \sin 2 x+C=-\frac{1}{2} \sin 2 x+C$

Comparing both sides, we get

$a=-\frac{1}{2}$