∫sec2x4tan2x+9dx 13tan−12tanx3+C 16tan−12tanx3+C 16tan−

$\int \frac{s e c^{2} (x)}{4 t a n^{2} (x) + 9} d x$

$\frac{1}{3} t a n^{- 1} (\frac{2 t a n (x)}{3}) + C$

$\frac{1}{6} t a n^{- 1} (\frac{2 t a n (x)}{3}) + C$

$\frac{1}{6} t a n^{- 1} (\frac{t a n (x)}{6}) + C$

$\frac{1}{6} t a n^{- 1} (\frac{2 t a n (x)}{5}) + C$

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

Here, if we put tan(x) = t the numerator $s e c^{2} (x)$ dx will become dt. As $s e c^{2} (x)$ 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, $s e c^{2} (x)$ dx = dt

And the given integral would be like

$\int \frac{1}{4 t^{2} + 9} d t O r \frac{1}{4} \int \frac{1}{t^{2} + \frac{9}{4}} d t O r \frac{1}{4} \int \frac{1}{t^{2} + 3 {(\frac{3}{2})}^{2}} d t$

We can see that this is of the form $\int \frac{1}{x^{2} + a^{2}} d x$

After using the corresponding formula and substituting back the value of “ t “ which is tan(x) we get the final answer equal to

$\frac{1}{6} t a n^{- 1} (\frac{2 t a n (x)}{3}) + C$

∫sec2x4tan2x+9dx 13tan−12tanx3+C 16tan−12tanx3+C 16tan−

Right Answer is: B

SOLUTION

Related QuestionsA continuous and differentiable function will have a un∫1x+1√x−2dx 2√3tan−1√x+1√3+C 2√3tan−1√x−2√3+C 2√3tan−1√Integration of an integrable function fx gives a family∫x132+x122 dx 7x43+37x73+2411x116+C 4x43+37x73+2411x116

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A continuous and differentiable function will have a un
∫1x+1√x−2dx 2√3tan−1√x+1√3+C 2√3tan−1√x−2√3+C 2√3tan−1√
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∫x132+x122 dx 7x43+37x73+2411x116+C 4x43+37x73+2411x116