Answer:
jshduudndnhdugdfyfjdjfuzuvsmvucndfyxndjgthf
cos²xdx
=
2
1
∫2cos
xdx using cos2x=2cos
x−1⇒2cos
x=cos2x+1
∫(cos2x+1)dx
[
sin2x
+x]+c
4
+
x
+c
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Answers & Comments
Answer:
jshduudndnhdugdfyfjdjfuzuvsmvucndfyxndjgthf
Answer:
cos²xdx
=
2
1
∫2cos
2
xdx using cos2x=2cos
2
x−1⇒2cos
2
x=cos2x+1
=
2
1
∫(cos2x+1)dx
=
2
1
[
2
sin2x
+x]+c
=
4
sin2x
+
2
x
+c