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Question : 103 of 120
Marks:
+1,
-0
Solution:
We have,
I=∫‌| (sin‌x+cos‌x)(2−sin‌2‌x) |
| sin22x |
dx Put
sin‌x−cos‌x=t⇒(sin‌x+cos‌x)dx=dt
and
(sin‌x−cos‌x)2=t2⇒1−sin‌2‌x=t2
⇒‌‌sin‌2‌x=1−t2 ∴‌‌I=∫‌| (2−(1−t2))dt |
| (1−t2)2 |
⇒‌‌I=∫‌ ⇒‌‌I=∫‌dt ⇒‌‌I=∫‌dt ⇒‌‌I=∫‌dt Put
t−‌=y⇒(1+‌)dt=dy ∴‌‌I=∫‌=−‌+C ⇒‌‌I=‌+C ⇒‌‌I=‌+C ⇒‌‌I=‌| sin‌x−cos‌x |
| sin‌2‌x |
+C
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