Question

\[Calculate: \int\limits_{}^{}\sin^2x*dx,\]
By rewriting the identity:\[\cos(2x)=1-2*\sin^2(x)\]

Answer 1

∫(sinx)^2 dx
= ∫ (1-cos2x)/2 dx
= (1/2) ∫ (1-cos2x) dx
= (1/2) [ x - (1/2) sin2x] +c
= (1/2)x - (1/4) sin2x +c

Answer 2

\[\sin^2x=\frac{1}{2}(1-\cos(2x))\]so the integral becomes\[\int\limits{\frac{1}{2}(1-\cos(2x))}dx\]\[=\int\limits{\frac{1}{2}-\frac{1}{2}\cos(2x)}dx\]\[=\int\limits{\frac{1}{2}dx }-\int\limits{\frac{1}{2}\cos(2x)dx}=\frac{1}{2}x-\frac{1}{4}\sin(2x)+C\]

Answer 3

Thanks! :)