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Question

14sin2x+9cos2x dx will be equal to -

A
13tan1(2tan(x)3)+C
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B
16tan1(2tan(x)3)+C
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C
16tan1(2tan(x)6)+C
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D
16tan1(2tan(x)5)+C
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Solution

The correct option is B 16tan1(2tan(x)3)+C
In solving the integration problem it is important to see the form in which integrand is. Here, the given integrand is in 1acos2x+bsin2x.
To solve such integrals we need to multiply numerator as well as denominator of the integrand by sec2x.
Let’s do that and see what we get.
sec2xdx4tan2x+9
Now, we know how to solve this. We have already solved these forms.
Here, if we put tan(x) = t the numerator sec2(x) dx will become dt, as sec2(x) is the derivative of tan(x). And we’ll be left with a quadratic equation in the denominator which we can solve.
Let’s substitute tan(x) = t
So, sec2(x) dx=dt
And the given integral would be like
14t2+9 dt
14 1t2+94 dt
14 1t2+(32)2 dt
We can see that this is of the form 1x2+a2 dx
So, we’ll use the corresponding formula which is
1x2+a2 dx=1atan1(xa)+C
So,we get 16tan1(2t3)+C
Let’s substitute value of "t" which is tan(x).
So, the final answer would be=16tan1(2tan(x)3)+C

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