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Question : 75 of 160
Marks:
+1 ,
-0
Solution:
Let,
I = ∫ ‌ 2 ‌ cos ‌ 3 ‌ x − 3 s i n ‌ 3 x cos ‌ 3 ‌ x + 2 s i n ‌ 3 x
‌ dx Put
3 x = t ⇒ 3 ‌ dx = d t ∴ ‌ ‌ I = ∫ ‌ 2 ‌ cos ‌ t − 3 s i n ‌ t cos ‌ t + 2 s i n ‌ t
( ‌ ) Let,
2 ‌ cos ‌ t − 3 s i n ‌ t = A ( cos ‌ t + 2 s i n ‌ t ) + ‌ B [ ‌ ( cos ‌ t + 2 s i n ‌ t ) ] ⇒ 2 ‌ cos ‌ t − 3 s i n ‌ t = A ( cos ‌ t + 2 s i n ‌ t ) ‌ + B ( − s i n ‌ t + 2 ‌ cos ‌ t ) ⇒ 2 = A + 2 B ‌ and ‌ − 3 = 2 A − B On solving these equations, we get
‌ A = − ‌ ‌ and ‌ B = ‌ ‌ ∴ I = ‌ [ ∫ ‌ − ( cos ‌ t + 2 s i n ‌ t ) + ‌ ( − s i n ‌ t + 2 ‌ cos ‌ t )
( cos ‌ t + 2 s i n ‌ t ] ‌ dt ] ‌ = ‌ [ − ‌ t + ‌ ‌ log | cos ‌ t + 2 s i n ‌ t | ] + C ‌ = ‌ ‌ log | cos ‌ 3 ‌ x + 2 s i n ‌ 3 x | − ‌ x + C ‌ [ ∵ t = 3 x ] © examsiri.com
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