Q.
If ∫3sinx+2cosx3cosx+2sinxdx=ax+bln2sinx+3cosx+C then
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a
a=−1213,b=1539
b
a=−713,b=613
c
a=1213,b=−1539
d
a=−713,b=−613
answer is C.
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Detailed Solution
Differentiating both sides, we get3sinx+2cosx3cosx+2sinx=a+b(2cosx−3sinx)(2sinx+3cosx)=sinx⋅(2a−3b)+cosx⋅(3a+2b)(3cosx+2sinx)Comparing like terms on both sides, we get 3=2a−3b,2=3a+2b⇒a=1213,b=−1539
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