Question

Integrate :-

Answer 1

@dlearner @Fifciol

Answer 2

\[\cos2x=\cos^2x-\sin^2x=1-2\sin^2x \rightarrow \sin^2x=\frac{ 1-\cos2x }{ 2 }\]

Answer 3

so
\[\int\limits_{}^{}\sin^2xdx=\int\limits \frac{dx }{ 2 }+\int\limits \frac{\cos2x }{ 2 }=\frac{ x }{ 2 }+\frac{ \sin2x }{ 4 }+C\]

Answer 4

should be - instead of +

Answer 5

great stuff @fifciol

Answer 6

thx

Answer 7

i wanted to ask where you got \[1-2\sin^2 x\]

or is it a trig identity law?

Answer 8

A law!!

Answer 9

\[1=\sin^2x+\cos^2x\]

Answer 10

well derived!!

Answer 11

and integral cos2x/2 by substitution u=2x

Answer 12

ohh ok i see now..thanks

Answer 13

Thank you