∫sinx dx/cos^2x√tan^2x-3
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Answered by
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Step-by-step explanation:
Explanation:
We have:
I
=
∫
π
3
0
sin
x
+
sin
x
tan
2
x
sec
2
x
d
x
=
∫
π
3
0
sin
x
+
sin
x
(
sin
2
x
cos
2
x
)
1
cos
2
x
d
x
=
∫
π
3
0
cos
2
x
(
sin
x
+
sin
3
x
cos
2
x
)
d
x
=
∫
π
3
0
(
cos
2
x
sin
x
+
sin
3
x
)
d
x
Let
sin
3
x
=
sin
2
x
sin
x
=
(
1
−
cos
2
x
)
sin
x
=
sin
x
−
cos
2
x
sin
x
:
I
=
∫
π
3
0
(
cos
2
x
sin
x
+
sin
x
−
cos
2
x
sin
x
)
d
x
=
∫
π
3
0
sin
x
d
x
Note that
∫
sin
x
d
x
=
−
cos
x
+
C
:
I
=
[
−
cos
x
]
π
3
0
=
−
cos
(
π
3
)
−
(
−
cos
(
0
)
)
=
−
1
2
−
(
−
1
)
=
1
2
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