What is $f (x) = \int {cos}^{2} x - {tan}^{3} x + sin x d x$ $f(x) = int cos^2x-tan^3x+sinx dx$ if $f (\frac{π}{6}) = 1$ $f(pi/6) = 1$ ?

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What is $f (x) = \int {cos}^{2} x - {tan}^{3} x + sin x d x$ $f(x) = int cos^2x-tan^3x+sinx dx$ if $f (\frac{π}{6}) = 1$ $f(pi/6) = 1$ ?

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Answer 1 · 2016-03-29T23:43:01

$f (x) = \frac{2 x + sin 2 x}{4} - \frac{{tan}^{2} x}{2} - ln | cos x | - cos x + \frac{28 - 2 π - 3 \sqrt{3}}{24} + ln ∣ ∣ ∣ \frac{\sqrt{3}}{2} ∣ ∣ ∣ + \frac{\sqrt{3}}{2}$ $f(x)=(2x+sin2x)/4-tan^2x/2-lnabs(cosx)-cosx+(28-2pi-3sqrt(3))/24+lnabs(sqrt(3)/2)+sqrt(3)/2$

Explanation:

Begin by using the sum rule to break $\int {cos}^{2} x - {tan}^{3} x + sin x d x$ $intcos^2x-tan^3x+sinxdx$ into:
$\int {cos}^{2} x d x - \int {tan}^{3} x d x + \int sin x d x$ $intcos^2xdx-inttan^3xdx+intsinxdx$

First Integral
Since we don't know what $\int {cos}^{2} x d x$ $intcos^2xdx$ is immediately, we have to tap into our trig knowledge. I hope you remember (or you can find, in your calculus textbook perhaps), the formula for ${cos}^{2} x$ $cos^2x$ :
${cos}^{2} x = \frac{1}{2} (1 + cos 2 x)$ $cos^2x=1/2(1+cos2x)$

Try to integrate this:
$\int {cos}^{2} x d x = \int \frac{1}{2} (1 + cos 2 x) d x$ $intcos^2xdx=int1/2(1+cos2x)dx$
$X X = \frac{1}{2} (\int 1 d x + \int cos 2 x d x)$ $color(white)(XX)=1/2(int1dx+intcos2xdx)$
$X X = \frac{1}{2} (x + C_{1} + \frac{sin 2 x}{2} + C_{2})$ $color(white)(XX)=1/2(x+C_1+(sin2x)/2+C_2)$
$X X = \frac{2 x + sin 2 x}{4} + C$ $color(white)(XX)=(2x+sin2x)/4+C$

Second Integral
Again, we don't know what $\int {tan}^{3} x d x$ $inttan^3xdx$ is, so we use our knowledge of trig (i.e. Pythagorean Identities) to solve it:
$\int {tan}^{3} x d x = \int tan x {tan}^{2} x d x$ $inttan^3xdx=inttanxtan^2xdx$
$X X = \int tan x ({sec}^{2} x - 1) d x$ $color(white)(XX)=inttanx(sec^2x-1)dx$
$X X = \int tan x {sec}^{2} x d x - \int tan x d x$ $color(white)(XX)=inttanxsec^2xdx-inttanxdx$
For the first of these, let $u = tan x$ $u=tanx$ and $\frac{d u}{d x} = {sec}^{2} x$ $(du)/dx=sec^2x$ ; that makes $d u = {sec}^{2} x d x$ $du=sec^2xdx$ :
$\int tan x {sec}^{2} x d x = \int u d u = \frac{u^{2}}{2} + C_{1}$ $inttanxsec^2xdx=intudu=u^2/2+C_1$
Because $u = tan x$ $u=tanx$ , $\int tan x {sec}^{2} x d x = \frac{tan x}{2} + C_{1}$ $inttanxsec^2xdx=(tanx)/2+C_1$ .

The second integral here, $\int tan x d x$ $inttanxdx$ , is simply $ln | cos x | + C_{2}$ $lnabs(cosx)+C_2$ . Combining these two results, we see $\int {tan}^{3} x = \frac{{tan}^{2} x}{2} + ln | cos x | + C$ $inttan^3x=tan^2x/2+lnabs(cosx)+C$ .

Third Integral
Thankfully, $\int sin x d x$ $intsinxdx$ is the most simple of all: it's just $- cos x + C$ $-cosx+C$ .

Now put all of it together:

$\int {cos}^{2} x d x = \frac{2 x + sin 2 x}{4} + C_{1}$ $intcos^2xdx=(2x+sin2x)/4+C_1$
$\int {tan}^{3} x d x = \frac{{tan}^{2} x}{2} + ln | cos x | + C_{2}$ $inttan^3xdx=tan^2x/2+lnabs(cosx)+C_2$
$\int sin x d x = - cos x + C_{3}$ $intsinxdx=-cosx+C_3$
$intcos^2xdx-inttan^3xdx+intsinxdx=(2x+sin2x)/4+C_1-tan^2x/2-lnabs(cosx)+C_2-cosx+C_3=(2x+sin2x)/4-tan^2x/2-lnabs(cosx)-cosx+C$

We are being asked for the value of $C$ if $f(pi/6)=1$ :
$1=(2(pi/6)+sin2(pi/6))/4-tan^2(pi/6)/2-lnabs(cos(pi/6))-cos(pi/6)+C$
$1=(pi/3+sqrt(3)/2)/4-(1/3)/2-lnabs(sqrt(3)/2)-sqrt(3)/2+C$
$1=(2pi+3sqrt(3))/24-1/6-lnabs(sqrt(3)/2)-sqrt(3)/2+C$
$1=(2pi+3sqrt(3)-4)/24-lnabs(sqrt(3)/2)-sqrt(3)/2+C$
$C=-(2pi+3sqrt(3)-4)/24+lnabs(sqrt(3)/2)+sqrt(3)/2+1$
$C=(28-2pi-3sqrt(3))/24+lnabs(sqrt(3)/2)+sqrt(3)/2$

Thus $f(x)=(2x+sin2x)/4-tan^2x/2-lnabs(cosx)-cosx+(28-2pi-3sqrt(3))/24+lnabs(sqrt(3)/2)+sqrt(3)/2$

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What is $f (x) = \int {cos}^{2} x - {tan}^{3} x + sin x d x$ $f(x) = int cos^2x-tan^3x+sinx dx$ if $f (\frac{π}{6}) = 1$ $f(pi/6) = 1$ ?

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What is $f (x) = \int {cos}^{2} x - {tan}^{3} x + sin x d x$ $f(x) = int cos^2x-tan^3x+sinx dx$ if $f (\frac{π}{6}) = 1$ $f(pi/6) = 1$ ?