Solve Indefinite Integral: 3 Ways

[lfdahl]

Solve the indefinite integral

\[\int \frac{dx}{\cos x+\sin x}\]

- in three different ways.

 

[topsquark]

I did it in W|A, my calculator, and I looked it up in a set of tables. Does that count? (Nerd)

-Dan

 

[lfdahl]

I did it in W|A, my calculator, and I looked it up in a set of tables. Does that count? (Nerd)

-Dan

Hi, topsquark/Dan!

Thankyou for your input.

You can use an online table or integrator to check your results, but the challenge is to solve the integral in three different ways without these means

 

[Greg]

Solve the indefinite integral

\[\int \frac{dx}{\cos x+\sin x}\]

- in three different ways.

Spoiler

1)

$$\int\frac{1}{\sin(x)+\cos(x)}\,dx=\int\frac{1}{\sqrt2\sin\left(x+\frac{\pi}{4}\right)}\,dx$$

$$=-\frac{1}{\sqrt2}\ln\left|\cot\left(x+\frac{\pi}{4}\right)+\csc\left(x+\frac{\pi}{4}\right)\right|+C$$

2) Unfortunately, this method of solution introduces a zero where the original integrand is defined. As an indefinite integral it is correct in the sense that its derivative is equivalent to the integrand, under certain circumstances.

$$\int\frac{1}{\cos(x)+\sin(x)}\,dx=\int\frac{\cos(x)-\sin(x)}{\cos(2x)}\,dx$$

$$=\int\frac{\cos(x)}{1-2\sin^2(x)}\,dx+\int\frac{\sin(x)}{1-2\cos^2(x)}\,dx$$

$$=\frac{1}{2\sqrt2}\left(\ln\left|1+\sqrt2\sin(x)\right|-\ln\left|1-\sqrt2\sin(x)\right|\right)$$

$$+\frac{1}{2\sqrt2}\left(\ln\left|1-\sqrt2\cos(x)\right|-\ln\left|1+\sqrt2\cos(x)\right|\right)+C$$

3)

$$\int\frac{1}{\sin(x)+\cos(x)}\,dx=\int\frac{1}{\frac{e^{ix}+e^{-ix}}{2}+\frac{e^{ix}-e^{-ix}}{2i}}\,dx$$

$$=\int\frac{2i}{(i+1)e^{ix}+(i-1)e^{-ix}}\,dx=\int\frac{2ie^{ix}}{(i+1)e^{2ix}+(i-1)}\,dx$$

$$\sqrt i\tan w=e^{ix},\quad\sqrt i\sec^2w\,dw=ie^{ix}\,dx$$

$$\sqrt2i\int\,dw=\sqrt2i w+C=-\sqrt2i\arctan\left(\frac{e^{ix}}{\sqrt i}\right)+C$$

$$=-\sqrt2i\arctan\left(e^{i(x-\pi/4)}\right)+C$$

(we need a negative to account for $i^2=-1$).

 

[lfdahl]

Spoiler

1)

$$\int\frac{1}{\sin(x)+\cos(x)}\,dx=\int\frac{1}{\sqrt2\sin\left(x+\frac{\pi}{4}\right)}\,dx$$

$$=-\frac{1}{\sqrt2}\ln\left|\cot\left(x+\frac{\pi}{4}\right)+\csc\left(x+\frac{\pi}{4}\right)\right|+C$$

2) Unfortunately, this method of solution introduces a zero where the original integrand is defined. As an indefinite integral it is correct in the sense that its derivative is equivalent to the integrand, under certain circumstances.

$$\int\frac{1}{\cos(x)+\sin(x)}\,dx=\int\frac{\cos(x)-\sin(x)}{\cos(2x)}\,dx$$

$$=\int\frac{\cos(x)}{1-2\sin^2(x)}\,dx+\int\frac{\sin(x)}{1-2\cos^2(x)}\,dx$$

$$=\frac{1}{2\sqrt2}\left(\ln\left|1+\sqrt2\sin(x)\right|-\ln\left|1-\sqrt2\sin(x)\right|\right)$$

$$+\frac{1}{2\sqrt2}\left(\ln\left|1-\sqrt2\cos(x)\right|-\ln\left|1+\sqrt2\cos(x)\right|\right)+C$$

3)

$$\int\frac{1}{\sin(x)+\cos(x)}\,dx=\int\frac{1}{\frac{e^{ix}+e^{-ix}}{2}+\frac{e^{ix}-e^{-ix}}{2i}}\,dx$$

$$=\int\frac{2i}{(i+1)e^{ix}+(i-1)e^{-ix}}\,dx=\int\frac{2ie^{ix}}{(i+1)e^{2ix}+(i-1)}\,dx$$

$$\sqrt i\tan w=e^{ix},\quad\sqrt i\sec^2w\,dw=ie^{ix}\,dx$$

$$\sqrt2i\int\,dw=\sqrt2i w+C=-\sqrt2i\arctan\left(\frac{e^{ix}}{\sqrt i}\right)+C$$

$$=-\sqrt2i\arctan\left(e^{i(x-\pi/4)}\right)+C$$

(we need a negative to account for $i^2=-1$).

Well done, greg1313! Thankyou for your participation!

Spoiler

It was your first two solutions, I had in mind. My 3rd solution is a tangent half angle substitution (Weierstrass)

 

[alyafey22]

The moment when the correct answer doesn't get most of the thanks.

 

[Greg]

(Giggle)

 

[lfdahl]

In case, anyone is interested in the tanget half angle solution, here it is:

Spoiler

$t = \tan \frac{x}{2}$: \[\int \frac{1}{\cos x + \sin x}dx = \int \frac{1}{\left ( \frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2} \right )}\frac{2dt}{1+t^2}= \int \frac{-2dt}{t^2-2t-1}=\int \frac{-2dt}{(t+\sqrt{2}-1)(t-\sqrt{2}-1)}= \\\\\frac{1}{\sqrt{2}}\int \left ( \frac{1}{t-1+\sqrt{2}}-\frac{1}{t-1-\sqrt{2}} \right )dt =\frac{1}{\sqrt{2}}\left ( \ln \left ( \left | t-1+\sqrt{2} \right | \right )-\ln \left ( \left | t-1-\sqrt{2} \right | \right ) \right )+C \\\\= \frac{1}{\sqrt{2}}\left ( \ln \left ( \left | \tan \frac{x}{2}-1+\sqrt{2} \right | \right )-\ln \left ( \left | \tan \frac{x}{2}-1-\sqrt{2} \right | \right ) \right )+C \\\\ = \frac{1}{\sqrt{2}} \ln \left ( \left | \frac{\tan \frac{x}{2}-1+\sqrt{2}}{\tan \frac{x}{2}-1-\sqrt{2}} \right | \right ) +C\]