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Question
`int(sin2x)/(5sin^2x+3cos^2x) dx=` ______.
Options
`1/2log abs(5sin^2x + 3cos^2x) +c`
`1/4log abs(3sin^2x + 5cos^2x) +c`
`1/4log abs(5sin^2x + 3cos^2x) +c`
`1/2log abs(3sin^2x + 5cos^2x) +c`
MCQ
Fill in the Blanks
Solution
`int(sin2x)/(5sin^2x+3cos^2x) dx= underline(1/2log abs(5sin^2x + 3cos^2x) +c)`.
Explanation:
Put 5sin2x + 3cos2x = t
⇒ (5 × 2 sinx cosx - 3 × 2 sinx cosx) dx = dt
⇒ 4 sinx cosx dx = dt
⇒ 2sin 2x dx = dt
`therefore int(sin2x)/(5sin^2x+3cos^2x) dx=1/2intdt/t`
`=1/2log abs(t)+c`
`=1/2log abs(5sin^2x + 3cos^2x) +c`
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