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Question : 74 of 160
Marks:
+1,
-0
Solution:
=∫‌| (1−4sin‌2x)‌cos‌x |
| cos(3x+2) |
‌dxLet
‌=∫‌| (1−4+4cos2x)‌cos‌x |
| cos(3x+2) |
‌dx=‌∫‌| 4cos2x−3‌cos‌x |
| cos(3x+2) |
‌dx=∫‌‌dxLet
3x+2=t⇒x‌dx=‌‌I=∫‌‌‌‌=‌‌∫‌| cos‌2‌cos‌t+sin‌2sin‌t |
| cos(t) |
‌dt‌‌=‌‌∫(cos‌2+sin‌2‌tan‌t)‌dt‌‌=‌[(cos‌2)⋅t+sin‌2‌ln( sect)+K]‌‌=‌[cos‌2⋅(3x+2)+sin‌2⋅ln[ sec(3x+2)]+K‌‌=(cos‌2)x+‌‌cos‌2+‌sin‌2⋅ln[ sec(3x+2)]+‌‌‌=(cos‌2)x+‌sin‌2‌ln[ sec(3x+2)]+C
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