integration of sin^2
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hai friend,
here is your answer
=∫sin²x dx
∫((1-cos2x)/2)dx
1/2 {∫dx -∫cos2x/2 dx}
1/2[x-sin2x/3]
x/2-sin2x/6
hope helped :)
here is your answer
=∫sin²x dx
∫((1-cos2x)/2)dx
1/2 {∫dx -∫cos2x/2 dx}
1/2[x-sin2x/3]
x/2-sin2x/6
hope helped :)
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