How do you evaluate the integral $\int \frac{{sin}^{2} x - {cos}^{2} x}{cos x}$ $int (sin^2x-cos^2x)/cosx$ from $[- \frac{π}{4}, \frac{π}{4}]$ $[-pi/4, pi/4]$ ?

Question

How do you evaluate the integral $\int \frac{{sin}^{2} x - {cos}^{2} x}{cos x}$ $int (sin^2x-cos^2x)/cosx$ from $[- \frac{π}{4}, \frac{π}{4}]$ $[-pi/4, pi/4]$ ?

1 Answer

Impact of this question

3479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-04T03:00:04

$\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{{sin}^{2} x - {cos}^{2} x}{cos x} d x = ln (3 + 2 \sqrt{2})$ $int_(-pi/4)^(pi/4) (sin^2x-cos^2x)/cosx dx = ln( 3+2sqrt2 )$

Explanation:

Using the fundamental trig Identity ${sin}^{2} A + {cos}^{2} A \equiv 1$ $sin^2A+cos^2A-=1$ we have

$\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{{sin}^{2} x - {cos}^{2} x}{cos x} d x = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1 - {cos}^{2} x - {cos}^{2} x}{cos x} d x$ $int_(-pi/4)^(pi/4) (sin^2x-cos^2x)/cosx dx = int_(-pi/4)^(pi/4) (1-cos^2x-cos^2x)/cosx dx$

$= \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1 - 2 {cos}^{2} x}{cos x} d x$ $= int_(-pi/4)^(pi/4) (1-2cos^2x)/cosx dx$
$= \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{cos x} - 2 cos x d x$ $= int_(-pi/4)^(pi/4) 1/cosx-2cosx dx$
$= \int_{- \frac{π}{4}}^{\frac{π}{4}} sec x - 2 cos x d x$ $= int_(-pi/4)^(pi/4) secx-2cosx dx$
$= {[ln | sec x + tan x | - 2 sin x]}_{- \frac{π}{4}}^{\frac{π}{4}}$ $= [ln|secx+tanx| -2sinx]_(-pi/4)^(pi/4)$

$= (ln ∣ ∣ sec (\frac{π}{4}) + tan (\frac{π}{4}) ∣ ∣ - 2 sin (\frac{π}{4})) - (ln ∣ ∣ sec (- \frac{π}{4}) + tan (- \frac{π}{4}) ∣ ∣ - 2 sin (- \frac{π}{4}))$ $= (ln|sec(pi/4)+tan(pi/4)| -2sin(pi/4)) - (ln|sec(-pi/4)+tan(-pi/4)| -2sin(-pi/4))$

$= (ln (\sqrt{2} + 1) - 2 \frac{1}{2} \sqrt{2}) - (ln (\sqrt{2} - 1) - 2 \frac{1}{2} \sqrt{2})$ $= (ln(sqrt2+1) -2 1/2 sqrt2) - (ln(sqrt2-1) -2 1/2 sqrt2)$
$= ln (\sqrt{2} + 1) - 2 \frac{1}{2} \sqrt{2} - ln (\sqrt{2} - 1) + 2 \frac{1}{2} \sqrt{2}$ $= ln(sqrt2+1) -2 1/2 sqrt2 - ln(sqrt2-1) +2 1/2 sqrt2$
$= ln (\sqrt{2} + 1) - ln (\sqrt{2} - 1)$ $= ln(sqrt2+1) - ln(sqrt2-1)$
$= ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})$ $= ln((sqrt2+1)/(sqrt2-1))$
$= ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1})$ $= ln( (sqrt2+1)/(sqrt2-1) * (sqrt2+1)/(sqrt2+1) )$
$= ln (\frac{2 + 2 \sqrt{2} + 1}{2 - 1})$ $= ln( (2+2sqrt2+1)/(2-1) )$
$= ln (3 + 2 \sqrt{2})$ $= ln( 3+2sqrt2 )$

Science

Math

Humanities

... and beyond

How do you evaluate the integral $\int \frac{{sin}^{2} x - {cos}^{2} x}{cos x}$ $int (sin^2x-cos^2x)/cosx$ from $[- \frac{π}{4}, \frac{π}{4}]$ $[-pi/4, pi/4]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate the integral ∫sin2x−cos2xcosxint (sin^2x-cos^2x)/cosx from [−π4,π4][-pi/4, pi/4]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you evaluate the integral $\int \frac{{sin}^{2} x - {cos}^{2} x}{cos x}$ $int (sin^2x-cos^2x)/cosx$ from $[- \frac{π}{4}, \frac{π}{4}]$ $[-pi/4, pi/4]$ ?