Question

integration of sin8x-cos8x​

Answer 1

Answer:

hey mate your answer is....

Step-by-step explanation:

Here use the formula, a^2-b^2=(a+b)(a-b), a^4+b^4=(a^2+b^2)-2a^2b^2

LHS is \sin^8 x- \cos ^8 x=(\sin^4 x- \cos ^4 x)(\sin^4 x+ \cos ^4 x)\\ \sin^8 x- \cos ^8 x=(\sin^2 x+ \cos ^2 x)(\sin^2 x- \cos ^2 x)[(\sin^2 x+ \cos ^2 x)^2-2\sin^2 x \cos ^2 x]\\ \sin^8 x- \cos ^8 x=(1)(\sin^2 x- \cos ^2 x)[(1)^2-2\sin^2 x \cos ^2 x]\\ \sin^8 x- \cos ^8 x=(\sin^2 x- \cos ^2 x)(1-2\sin^2 x \cos ^2 x)\\ =RHS

@rajsingh....

Answer 2

Answer:

integration of sin8x-cos8x

Step-by-step explanation:

(sin8x=cos8x/8

(cos8x=-sin8x/8

cos8x/8+sin8x/8