# How do you find int5/(16+9cos^2x)dx?

Jul 22, 2018

$I = \frac{1}{4} a r c \tan \left(\frac{4}{5} \tan x\right) + c$

#### Explanation:

Here ,

$I = \int \frac{5}{16 + 9 {\cos}^{2} x} \mathrm{dx}$

$= \int \frac{5 {\sec}^{2} x}{16 {\sec}^{2} x + 9} \mathrm{dx} \ldots \ldots \ldots \to \left[\because \cos \theta = \frac{1}{\sec} \theta\right]$

$= \int \frac{5 {\sec}^{2} x}{16 \left(1 + {\tan}^{2} x\right) + 9} \mathrm{dx} \to \left[\because {\sec}^{2} \theta - {\tan}^{2} \theta = 1\right]$

$= 5 \int {\sec}^{2} \frac{x}{16 {\tan}^{2} x + 25} \mathrm{dx}$

Subst. color(blue)(tanx=u=>sec^2xdx=du

$\therefore I = 5 \int \frac{1}{16 {u}^{2} + 25} \mathrm{du}$

$= \frac{5}{16} \int \frac{1}{{u}^{2} + \frac{25}{16}} \mathrm{du}$

$= \frac{5}{16} \int \frac{1}{{u}^{2} + {\left(\frac{5}{4}\right)}^{2}} \mathrm{du}$

$= \frac{5}{16} \cdot \frac{1}{\frac{5}{4}} \arctan \left(\frac{u}{\frac{5}{4}}\right) + c$

$= \frac{5}{16} \times \frac{4}{5} a r c \tan \left(\frac{4 u}{5}\right) + c$

$\therefore I = \frac{1}{4} \arctan \left(\frac{4 u}{5}\right) + c$

Subst. back ,color(blue)(u=tanx

$I = \frac{1}{4} a r c \tan \left(\frac{4}{5} \tan x\right) + c$

Jul 22, 2018

$\frac{1}{4} \arctan \left(\frac{4}{5} \tan x\right) + C$.

#### Explanation:

Let, $I = \int \frac{5}{16 + 9 {\cos}^{2} x} \mathrm{dx}$.

:. I=5int1/{cos^2x(16/cos^2x+9)dx,

$= 5 \int \left\{\left(\frac{1}{\cos} ^ 2 x\right) \left(\frac{1}{16 {\sec}^{2} x + 9}\right)\right\} \mathrm{dx}$,

$= 5 \int \left\{\frac{1}{16 \left({\tan}^{2} x + 1\right) + 9}\right\} {\sec}^{2} x \mathrm{dx}$.

This suggests that the substn. $4 \tan x = u$ must work.

Now, $4 \tan x = u \Rightarrow {\sec}^{2} x \mathrm{dx} = \frac{1}{4} \mathrm{du}$.

$\therefore I = 5 \int \left\{\frac{1}{{u}^{2} + {5}^{2}}\right\} \frac{1}{4} \mathrm{du}$,

$= \frac{5}{4} \cdot \frac{1}{5} \arctan \left(\frac{u}{5}\right)$.

$\Rightarrow I = \frac{1}{4} \arctan \left(\frac{4}{5} \tan x\right) + C$.