Question

Find value of $\int {\dfrac{{dx}}{{9 + 16{{\sin }^2}x}}}$ 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

We are given an integral and first using the identity ${\cos ^2}x + {\sin ^2}x = 1$ we get $\int {\dfrac{{dx}}{{9{{\cos }^2}x + 25{{\sin }^2}x}}}$ and further taking ${\cos ^2}x$ common in the denominator and simplifying we a new integrand and using the method of substitution we can solve the integral which is of the form $\int {\dfrac{{dx}}{{{a^2} + {x^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + c$

Complete step-by-step answer:
Here we need to find the value of $\int {\dfrac{{dx}}{{9 + 16{{\sin }^2}x}}}$
We know that ${\cos ^2}x + {\sin ^2}x = 1$
Implying this we get
$\Rightarrow \int {\dfrac{{dx}}{{9\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + 16{{\sin }^2}x}}} \\\ \Rightarrow \int {\dfrac{{dx}}{{9{{\cos }^2}x + 9{{\sin }^2}x + 16{{\sin }^2}x}}} \\\ \Rightarrow \int {\dfrac{{dx}}{{9{{\cos }^2}x + 25{{\sin }^2}x}}} \\\$
Now lets take ${\cos ^2}x$ common in the denominator
$\Rightarrow \int {\dfrac{{dx}}{{{{\cos }^2}x\left( {9 + \dfrac{{25{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}} \\\ \Rightarrow \int {\dfrac{{{{\sec }^2}xdx}}{{\left( {9 + 25{{\tan }^2}x} \right)}}} \\\$
The above is given by $\dfrac{1}{{{{\cos }^2}x}} = {\sec ^2}x$ and $\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x$
Let it be equation (1)
Now let's solve this by substitution method
Let's take tan x = t
$\Rightarrow {\sec ^2}xdx = dt$
Applying this in equation (1)
$\Rightarrow \int {\dfrac{{dt}}{{\left( {9 + 25{\operatorname{t} ^2}} \right)}}} \\\ \Rightarrow \int {\dfrac{{dt}}{{\left( {{3^2} + {{(5\operatorname{t} )}^2}} \right)}}} \\\ \\\$
This is of the form $\int {\dfrac{{dx}}{{{a^2} + {x^2}}}}$
We know that $\int {\dfrac{{dx}}{{{a^2} + {x^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + c$
Here a = 3 and x = 5t
Therefore
$\Rightarrow \int {\dfrac{{dt}}{{\left( {{3^2} + {{(5\operatorname{t} )}^2}} \right)}}} = \dfrac{1}{3}{\tan ^{ - 1}}\dfrac{{5t}}{3} + c \\\ \\\$
Substituting t = tan x
$\Rightarrow \int {\dfrac{{dt}}{{\left( {{3^2} + {{(5\operatorname{t} )}^2}} \right)}}} = \dfrac{1}{3}{\tan ^{ - 1}}\left( {\dfrac{{5\tan x}}{3}} \right) + c \\\ \\\$
None of the options are correct.

Additional information: Integrals are used extensively in many areas of mathematics as well as in many other areas that rely on mathematics. For example, in probability theory, integrals are used to determine the probability of some random variable falling within a certain range.
An integral is the reverse of a derivative. A derivative is the steepness (or "slope"), as the rate of change, of a curve. The word "integral" can also be used as an adjective meaning "related to integers”.

Note: Try to remember basic trigonometric identities and integration formula. Using trigonometric formulas reduces the equation and solve further to get an answer.