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Question : 76 of 89
Marks:
+1,
-0
Solution:
∫‌‌dx=∫‌‌dx=∫(‌×‌| cos‌x |
| sin‌x−2‌cos‌x |
)‌dx=∫‌| 5sin‌x‌dx |
| sin‌x−2‌cos‌x |
=∫(‌| 4sin‌x+sin‌x+2‌cos‌x−2‌cos‌x |
| sin‌x−2‌cos‌x |
)‌dx=∫‌| (sin‌x−2‌cos‌x)+(4sin‌x+2‌cos‌x) |
| sin‌x−2‌cos‌x |
‌dx=∫‌| (sin‌x−2‌cos‌x)+2(cos‌x+2sin‌x) |
| (sin‌x−2‌cos‌x) |
‌dx=∫‌| sin‌x−2‌cos‌x |
| sin‌x−2‌cos‌x |
‌dx+2‌∫(‌| cos‌x+2sin‌x |
| sin‌x−2‌cos‌x |
)‌dx =∫dx+2‌∫‌| cos‌x+2sin‌x |
| sin‌x−2‌cos‌x |
‌dx=I1+I2 where
I1=∫dx and
I2=2‌∫‌| cos‌x+2sin‌x |
| sin‌x−2‌cos‌x |
‌dx Put
sin‌x−2‌cos‌x=t‌⇒(cos‌x+2sin‌x)‌dx=dt‌∴I2=2‌∫‌=2‌ln‌t+C‌=2‌ln(sin‌x−2‌cos‌x)+CHence,
I1+I2‌=∫dx+2‌ln(sin‌x−2‌cos‌x)+c‌=x+2‌ln|(sin‌x−2‌cos‌x)|+k‌⇒a=2
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