32 sin 4x cos 4x dx. Find value of integration
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Explanation:
I=∫
sin
4
x+cos
4
x
1
dx
=∫
1+tan
4
x
sec
4
x
dx
Let tanx=t
⇒sec
2
xdx=dt
I=∫
1+t
4
(1+t
2
)
dt
=∫
t
2
(t
2
+
t
2
1
)
t
2
(1+
t
2
1
)
dt
=∫
t
2
+
t
2
1
−2+2
1+
t
2
1
dt
Let t−
t
1
=v
⇒(1+
t
2
1
)dt=dv
I=∫
(
2
)
2
+v
2
dv
=
2
1
tan
−1
(
2
v
)+c
=
2
1
tan
−1
[
2
tanx
tan
2
x+1
]+c.
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