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Question

Evaluate sec2(3x+4)dx

A
tan(3x+4)3+C
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B
tan(3x+4)4+C
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C
tan(3x+4)3+C
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D
cot(3x+4)4+C
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Solution

The correct option is A tan(3x+4)3+C
As we know that f(ax+b)dx=F(ax+b)a+C, where a,b and C are constants.
We have here,
f(x)=sec2(3x+4), where a=3,b=4
So, sec2(3x+4)dx=tan(3x+4)3+C

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