How do you integrate $\int \frac{{sec}^{2} x}{{(4 - {tan}^{2} x)}^{\frac{3}{2}}}$ $int sec^2x/(4-tan^2x)^(3/2)$ by trigonometric substitution?

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How do you integrate $\int \frac{{sec}^{2} x}{{(4 - {tan}^{2} x)}^{\frac{3}{2}}}$ $int sec^2x/(4-tan^2x)^(3/2)$ by trigonometric substitution?

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Answer 1 · 2016-09-11T15:12:08

$\frac{tan x}{4 \sqrt{4 - {tan}^{2} x}} + C$ $tanx/(4sqrt(4-tan^2x))+C$

Explanation:

We have:

$I = \int \frac{{sec}^{2} x}{{(4 - {tan}^{2} x)}^{\frac{3}{2}}} d x$ $I=intsec^2x/(4-tan^2x)^(3/2)dx$

We will first use the non-trigonometric substitution $u = tan x$ $u=tanx$ , implying that $d u = {sec}^{2} x d x$ $du=sec^2xdx$ :

$I = \int \frac{d u}{{(4 - u^{2})}^{\frac{3}{2}}}$ $I=int(du)/(4-u^2)^(3/2)$

Now we will apply the trigonometric substitution $u = 2 sin θ$ $u=2sintheta$ . Recall that this implies that $d u = 2 cos θ d θ$ $du=2costhetad theta$ .

$I = \int \frac{2 cos θ d θ}{{(4 - 4 {sin}^{2} θ)}^{\frac{3}{2}}}$ $I=int(2costhetad theta)/(4-4sin^2theta)^(3/2)$

$= \int \frac{2 cos θ d θ}{4^{\frac{3}{2}} {(1 - {sin}^{2} θ)}^{\frac{3}{2}}}$ $=int(2costhetad theta)/(4^(3/2)(1-sin^2theta)^(3/2))$

Note that $1 - {sin}^{2} θ = {cos}^{2} θ$ $1-sin^2theta=cos^2theta$ :

$I = \int \frac{2 cos θ d θ}{8 {({cos}^{2} θ)}^{\frac{3}{2}}}$ $I=int(2costhetad theta)/(8(cos^2theta)^(3/2))$

$= \int \frac{cos θ d θ}{4 {cos}^{3} θ}$ $=int(costhetad theta)/(4cos^3theta)$

$= \frac{1}{4} \int \frac{d θ}{{cos}^{2} θ}$ $=1/4int(d theta)/cos^2theta$

$= \frac{1}{4} \int {sec}^{2} θ d θ$ $=1/4intsec^2thetad theta$

$= \frac{1}{4} tan θ + C$ $=1/4tantheta+C$

Recall that $u = 2 sin θ$ $u=2sintheta$ , so $θ = arcsin (\frac{u}{2})$ $theta=arcsin(u/2)$ .

$I = \frac{1}{4} tan (arcsin (\frac{u}{2})) + C$ $I=1/4tan(arcsin(u/2))+C$

We can simplify this: draw the triangle where sine is $\frac{u}{2}$ $u/2$ , that is, the opposite side is $u$ $u$ and the hypotenuse is $2$ $2$ . Through the Pythagorean theorem, we see that the adjacent side is $\sqrt{4 - u^{2}}$ $sqrt(4-u^2)$ .

Thus, the tangent is opposite over adjacent, or:

$I = \frac{1}{4} (\frac{u}{\sqrt{4 - u^{2}}}) + C$ $I=1/4(u/sqrt(4-u^2))+C$

Now, since $u = tan x$ $u=tanx$ :

$I = \frac{tan x}{4 \sqrt{4 - {tan}^{2} x}} + C$ $I=tanx/(4sqrt(4-tan^2x))+C$

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How do you integrate $\int \frac{{sec}^{2} x}{{(4 - {tan}^{2} x)}^{\frac{3}{2}}}$ $int sec^2x/(4-tan^2x)^(3/2)$ by trigonometric substitution?

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How do you integrate $\int \frac{{sec}^{2} x}{{(4 - {tan}^{2} x)}^{\frac{3}{2}}}$ $int sec^2x/(4-tan^2x)^(3/2)$ by trigonometric substitution?