Question

Solve the given integration to choose the correct answer :
$\int {\dfrac{{dx}}{{9 + 16{{\sin }^2}x}}}$ is equal to
A. $\dfrac{1}{3}{\tan ^{ - 1}}\left( {\dfrac{{3\tan x}}{5}} \right) + c$.
B. $\dfrac{1}{5}{\tan ^{ - 1}}\left( {\dfrac{{\tan x}}{{15}}} \right) + c$.
C. $\dfrac{1}{{15}}{\tan ^{ - 1}}\left( {\dfrac{{\tan x}}{5}} \right) + c$.
D. $\dfrac{1}{{15}}{\tan ^{ - 1}}\left( {\dfrac{{5\tan x}}{3}} \right) + c$.

Answer 1

Hint : In this question, we will use the algorithm to evaluate the different forms of integral and also use the method of integration by substitution. The form $\int {\dfrac{1}{{a + b{{\sin }^2}x}}dx}$ can be evaluated by using the following algorithm.
Step 1 : divide numerator and denominator both by ${\cos ^2}x$.
Step 2 : replace ${\sec ^2}x$, if any, in denominator by $1 + {\tan ^2}x$.
Step 3 : put tan x = t so that ${\sec ^2}xdx = dt$. This substitution reduces the integral in the form $\int {\dfrac{1}{{a{t^2} + bt + c}}dt}$.
Step 4 : evaluate the integral obtained in step 3 by using the suitable methods.

Complete step-by-step answer:
The given integral is,
$\int {\dfrac{{dx}}{{9 + 16{{\sin }^2}x}}}$
First, Divide the numerator and denominator both by ${\cos ^2}x$.
$\Rightarrow \int {\dfrac{{\dfrac{1}{{{{\cos }^2}x}}dx}}{{\dfrac{{9 + 16{{\sin }^2}x}}{{{{\cos }^2}x}}}}} = \int {\dfrac{{{{\sec }^2}xdx}}{{9{{\sec }^2}x + 16{{\tan }^2}x}}}$
Now, Replace ${\sec ^2}x$ by $1 + {\tan ^2}x$.
$\Rightarrow \int {\dfrac{{{{\sec }^2}xdx}}{{9(1 + {{\tan }^2}x) + 16{{\tan }^2}x}}} \\\ \Rightarrow \int {\dfrac{{{{\sec }^2}xdx}}{{9 + 25{{\tan }^2}x}}} \\\$
$\Rightarrow \int {\dfrac{{{{\sec }^2}xdx}}{{{{(3)}^2} + {{(5\tan x)}^2}}}}$ ……….. (i)
Let t = 5 tan x
Differentiating the both sides, we will get
$dt = 5{\sec ^2}xdx \\\ \dfrac{{dt}}{5} = {\sec ^2}xdx \\\$
Putting this value in equation (i), we get
$\Rightarrow \dfrac{1}{5}\int {\dfrac{{dt}}{{{{(3)}^2} + {{(t)}^2}}}}$. ……… (ii)
As we know that
$\int {\dfrac{{dx}}{{{{(x)}^2} + {{(a)}^2}}} = \dfrac{1}{a}{{\tan }^{ - 1}}\left( {\dfrac{x}{a}} \right) + c}$.
Thus, comparing it with equation (ii), it will become,
$\Rightarrow \dfrac{1}{5}\int {\dfrac{{dt}}{{{{(3)}^2} + {{(t)}^2}}} = \dfrac{1}{{3 \times 5}}{{\tan }^{ - 1}}\left( {\dfrac{t}{3}} \right) + c} \\\ \Rightarrow \dfrac{1}{{15}}{\tan ^{ - 1}}\left( {\dfrac{t}{3}} \right) + c \\\$
Now, we will replace the value of t by 5 tan x.
Then we get,
$\Rightarrow \dfrac{1}{{15}}{\tan ^{ - 1}}\left( {\dfrac{{5\tan x}}{3}} \right) + c$
Hence, we can say that $\int {\dfrac{{dx}}{{9 + 16{{\sin }^2}x}}}$ is equal to $\dfrac{1}{{15}}{\tan ^{ - 1}}\left( {\dfrac{{5\tan x}}{3}} \right) + c$.
Therefore, the correct answer is option (D).

Note : Whenever we ask such types of questions, we will use the methods of solving the different integral forms. First, we have to simplify the given integral using the suitable algorithm according to its form. Then we will evaluate that obtained integral step by step by using the substitution method. After that we can easily solve that and through this, we will get the required answer.