What is $intsin^2(x) dx$ ?

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Answer 1 · 2014-12-15T15:03:02

We can solve this by expressing $*sin ^2 (x)$ in terms of a function in the first power.

From the equation:

$cos 2x = cos ^ 2 (x) - sin^2 (x)$

and the equation

$sin^2 (x) + cos^2 (x) = 1$

$cos ^2 (x) = 1 - sin^2 (x)$

then. we substitute to the first equation

$cos 2x = 1 - sin^2 (x) - sin ^2(x)$

simplifying,

$sin ^2 (x) = (1 - cos 2x)/ 2$

substitute to the original problem

$int [(1-cos 2x)/2]dx$

$(1/2)int(1 - cos 2x)dx$

$(1/2)[ intdx - intcos 2xdx$

$(1/2)[x - sin 2x*intd(2x)]$

$(1/2)[x - (sin 2x) * 2*1]$

$x/2 - sin 2x + c$

What is intsin^2(x) dxintsin^2(x) dx?