show that (sin+cos)²-(sin-cos)^2?=4 sin×
cos
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Answered by
0
AnswerWe have,
I=∫
sin
2
2x
(sinx+cosx)(2−sin2x)
dx
Put sinx−cosx=1⇒(sinx+cosx)dx=dt and (sinx−cosx)
2
=t
2
⇒1−sin2x=t
2
⇒sin2x=1−t
2
∴I=∫
(1−t
2
)
2
(2−(1−t
2
))dt
⇒I=∫
(1−t
2
)
2
(1+t
2
)dt
⇒I=∫
1−2t
2
+t
4
1+t
2
dt
⇒I=∫
t
2
1
+t
2
−2
1+
t
2
1
dt
⇒=∫
(t−
t
1
)
2
1+
t
2
1
dt
Put t−
t
1
=y⇒(1+
t
2
1
)dt=dy
∴I=∫
y
2
dy
=−
y
1
+C
⇒I=
t−
t
1
−1
+C
⇒I=
1−t
2
t
+C
⇒I=
sin2x
sinx−cosx
+C
Step-by-step explanation:
HOPE IT HELPS YOU!!
Answered by
1
Answer:
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