# What is int (sin^2x)/(cos^3(x)) dx?

Jun 18, 2018

$\frac{1}{2} \sec x \tan x - \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid + C$

#### Explanation:

$\int {\sin}^{2} \frac{x}{\cos} ^ 3 x \mathrm{dx} = \int \frac{1 - {\cos}^{2} x}{\cos} ^ 3 x \mathrm{dx} = \int \left({\sec}^{3} x - \sec x\right) \mathrm{dx}$

Both of these are fairly tricky. Here's a link that should help out a lot: https://www.math.ubc.ca/~feldman/m121/secx.pdf

Using these results, we get an answer:

$= \left(\frac{1}{2} \sec x \tan x + \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid\right) - \ln \left\mid \sec x + \tan x \right\mid + C$

$= \frac{1}{2} \sec x \tan x - \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid + C$

Jun 18, 2018

 1/2[secxtanx-ln|(secx+tanx)|+C,

$\mathmr{and} , \frac{1}{2} \left[\sin \frac{x}{\cos} ^ 2 x - \ln | \frac{1 + \sin x}{\cos} x |\right] + C$.

#### Explanation:

Let, $I = \int {\sin}^{2} \frac{x}{\cos} ^ 3 x \mathrm{dx} = \int \sin \frac{x}{\cos} x \cdot \frac{1}{\cos} x \cdot \sin \frac{x}{\cos} x \mathrm{dx}$,

$= \int \left(\tan x\right) \left(\sec x \tan x\right) \mathrm{dx}$.

So, letting, $\sec x = t , \text{ we have, } \sec x \tan x \mathrm{dx} = \mathrm{dt}$.

Also, $\tan x = \sqrt{{\sec}^{2} x - 1}$.

$\therefore I = \int \sqrt{{t}^{2} - 1} \mathrm{dt}$,

$= \frac{1}{2} \left[t \sqrt{{t}^{2} - 1} - \ln | \left(t + \sqrt{{t}^{2} - 1}\right) |\right]$.

Since, $t = \sec x$, we get,

 I=1/2[secxtanx-ln|(secx+tanx)|+C,

$\mathmr{and} , I = \frac{1}{2} \left[\sin \frac{x}{\cos} ^ 2 x - \ln | \frac{1 + \sin x}{\cos} x |\right] + C$.