integration sin⁷x dx
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Answered by
7
I =sin^7x.dx
=sin^6x .sinx.dx
= (sin²x)³.sinx.dx
=( 1 - cos²x)³sinx.dx
={1 -3cos²x +3cos⁴x - cos^6x}sinx.dx
=sinx.dx -3cos²xsinx.dx + 3cos⁴x.sinx.dx -cos^6x.sinx.dx
= -cosx + {3cos²x( -sinx).dx } -3/5{5cos⁴x(-sinx)dx +1/7{7cos^6x.(-sinx).} dx
[ you know , I think
I = f(x)^n.f(x)'.dx = f(x)^(n +1)/( n +1) ]
= - cosx +(cosx)³ -3/5(cosx)^5 + 1/7(cosx)^7 + Constant
=sin^6x .sinx.dx
= (sin²x)³.sinx.dx
=( 1 - cos²x)³sinx.dx
={1 -3cos²x +3cos⁴x - cos^6x}sinx.dx
=sinx.dx -3cos²xsinx.dx + 3cos⁴x.sinx.dx -cos^6x.sinx.dx
= -cosx + {3cos²x( -sinx).dx } -3/5{5cos⁴x(-sinx)dx +1/7{7cos^6x.(-sinx).} dx
[ you know , I think
I = f(x)^n.f(x)'.dx = f(x)^(n +1)/( n +1) ]
= - cosx +(cosx)³ -3/5(cosx)^5 + 1/7(cosx)^7 + Constant
Anonymous:
yep :(
I = sinx.dx -3cos²xsinx.dx +3cos⁴x.sinxdx -cos^6x.sinxdx
you know sinx.dx = -cosx
yep that I know
i don't wanna hurt u but i am really confusing :(
-3cos²x.sinx .dx = d(cos³x)/dx .dx
where you confused
tell me which step
5th step
( a - b)³ = a³ - 3a²b+ 3ab² - b³
Answered by
6
Here's the solution. Hope it helps.
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