^8x-^8x/1-2²x²x - NCERT Class XII Mathematics Chapter -

NCERT Class XII Mathematics Chapter - - Solutions

..... Miscellaneous Exercise .....
Integrate the functions in Exercises 229 to 252.

Question : 238 of 271

Marks: +1, -0

\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}

Solution:

Let

I=\int\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}dx

We have,

(\sin^8x-\cos^8x)=(\sin^4x+\cos^4x)(\sin^4x-\cos^4x)

=[(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x](\sin^2x+\cos^2x)(\sin^2x-\cos^2x)

=(1-2\sin^2x\cos^2x)(1)(-\cos2x)

\therefore I=\int\frac{(1-2\sin^2x\cos^2x)(1)(-\cos2x)}{1-2\sin^2x\cos^2x}dx

=-\int\cos2xdx

=-\frac{1}{2}\sin2x+C

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