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

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

Answers (1)

Answer:

-\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}

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