    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 16tan−1(2tan(x)3)+CIn 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=1atan−1(xa)+C So,we get 16tan−1(2t3)+C Let’s substitute value of "t" which is tan(x). So, the final answer would be=16tan−1(2tan(x)3)+C  Suggest Corrections  0      Similar questions  Related Videos   Integration of Trigonometric Functions
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