How do you find $\int \frac{{sec}^{2} x}{1 - {sin}^{2} x} d x$ $int sec^2x/(1-sin^2x) dx$ ?

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How do you find $\int \frac{{sec}^{2} x}{1 - {sin}^{2} x} d x$ $int sec^2x/(1-sin^2x) dx$ ?

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Answer 1 · 2018-05-26T14:20:41

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Explanation:

${sin}^{2} x + {cos}^{2} x = 1 \Rightarrow {cos}^{2} x = 1 - {sin}^{2} x$ $sin^2 x + cos^2 x = 1 => cos^2x = 1 - sin^2 x$

$\Rightarrow \int \frac{{sec}^{2} x}{{cos}^{2} x} d x$ $=> int sec^2x / ( cos^2 x ) dx$

$\Rightarrow \int {sec}^{2} x \cdot \frac{1}{{cos}^{2} x} d x$ $=> int sec^2 x * 1/cos^2 x dx$

$\Rightarrow \int {sec}^{4} x d x$ $=> int sec^4 x dx$

$\Rightarrow \int {sec}^{2} x \cdot {sec}^{2} x d x$ $=> int sec^2 x * sec^2 x dx$

$\Rightarrow \int (1 + {tan}^{2} x) \cdot {sec}^{2} x d x$ $=> int ( 1 + tan^2x) * sec^2 x dx$

$u = tan x$ $u = tanx$

$d u = {sec}^{2} x d x$ $du = sec^2 x dx$

$\Rightarrow \int 1 + u^{2} d u$ $=> int 1 + u ^2 du$

$\Rightarrow u + \frac{1}{3} u^{3} + c$ $=> u + 1/3 u^3 + c$

$= tan x + \frac{1}{3} {tan}^{3} x + c$ $= tanx + 1/3 tan^3 x + c$

Answer 2 · 2018-05-26T14:28:16

$tan (x) + \frac{1}{3} {tan}^{3} (x) + C$ $tan(x) + frac(1)(3) tan^(3)(x) + C$

Explanation:

We have: $\int$ $int$ $\frac{{sec}^{2} (x)}{1 - {sin}^{2} (x)}$ $frac(sec^(2)(x))(1 - sin^(2)(x))$ $d x$ $dx$

$= \int$ $= int$ $\frac{{sec}^{2} (x)}{{cos}^{2} (x)}$ $frac(sec^(2)(x))(cos^(2)(x))$ $d x$ $dx$

$= \int$ $= int$ $\frac{{sec}^{2} (x)}{\frac{1}{{sec}^{2} (x)}}$ $frac(sec^(2)(x))(frac(1)(sec^(2)(x)))$ $d x$ $dx$

$= \int$ $= int$ ${sec}^{2} (x) \cdot {sec}^{2} (x)$ $sec^(2)(x) cdot sec^(2)(x)$ $d x$ $dx$

Then, the Pythagorean identity is ${cos}^{2} (x) + {sin}^{2} (x) = 1$ $cos^(2)(x) + sin^(2)(x) = 1$ .

We can divide through by ${cos}^{2} (x)$ $cos^(2)(x)$ it to get:

$\Rightarrow 1 + {tan}^{2} (x) = {sec}^{2} (x)$ $Rightarrow 1 + tan^(2)(x) = sec^(2)(x)$

Let's apply this rearranged identity to our integral:

$= \int$ $= int$ $(1 + {tan}^{2} (x)) \cdot {sec}^{2} (x)$ $(1 + tan^(2)(x)) cdot sec^(2)(x)$ $d x$ $dx$

Now, let's use $u$ $u$ -substitution, where $u = tan (x) \Rightarrow d u = {sec}^{2} (x)$ $u = tan(x) Rightarrow du = sec^(2)(x)$ $d x$ $dx$ :

$= \int$ $= int$ $(1 + u^{2})$ $(1 + u^(2))$ $d u$ $du$

$= \int$ $= int$ $1$ $1$ $d u + \int$ $du + int$ $u^{2}$ $u^(2)$ $d u$ $du$

$= u + \frac{1}{3} u^{3} + C$ $= u + frac(1)(3) u^(3) + C$

Finally, we can substitute $tan (x)$ $tan(x)$ in place of $u$ $u$ :

$= tan (x) + \frac{1}{3} {tan}^{3} (x) + C$ $= tan(x) + frac(1)(3) tan^(3)(x) + C$

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2 Answers

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