More Craziness from Schaum's

[ninguen]

I did this one in Mathway, and it said it couldn't be done. And yet Schaum's insists that this integral can be evaluated... and even gives you a hint!

∫ 1/(1+cos(x) dx

should evaluate to

-cotx+ cscx + C

Mathway says it's impossible.

Schaum's insists it IS possible, but you have to multiply the numerator and denominator by 1-cos(x).

This gives me

∫(1-cosx)/sin^2(x) dx

I plug it into Mathway, and they still insists it cannot be found.

Any ideas?

 

[galactus]

\(\displaystyle \displaystyle \int\frac{1}{1+cos(x)}dx\)

Yes, it is integrable. Regardless of what Mathway says. Schaum is correct. But, here is another way other than their hint.

Rewrite as:

\(\displaystyle \displaystyle \int\frac{1}{\underbrace{cos^{2}(x/2)+sin^{2}(x/2)}_{\text{1}}+\underbrace{cos^{2}(x/2)-sin^{2}(x/2)}_{\text{cos(x)}}}dx\)

\(\displaystyle \displaystyle \frac{1}{2}\int \frac{1}{cos^{2}(x/2)}dx\)

\(\displaystyle =\displaystyle \frac{1}{2}\int sec^{2}(x/2)dx\)

Can you finish now?.

 

[pka]

∫ 1/(1+cos(x) dx
should evaluate to
-cotx+ cscx + C

\(\displaystyle \dfrac{1-\cos(x)}{\sin^2(x)}=\csc^2(x)-\dfrac{\cos(x)}{\sin^2(x)}\).

\(\displaystyle \displaystyle - \int {\frac{{\cos (x)}}{{\sin ^2 (x)}}dx = } \frac{1}{{\sin (x)}} = \csc (x)\)

P.S. Have you found this resource?

 

[soroban]

Hello, ninguen!

I did this one in Mathway, and it said it couldn't be done.
And yet Schaum's insists that this integral can be evaluated... and even gives you a hint!

\(\displaystyle \displaystyle \int \frac{dx}{1+\cos x}\)

should evaluate to: .\(\displaystyle -\cot x+ \csc x + C\)

Mathway says it's impossible. . . . . It's wrong!

\(\displaystyle \text{Multiply by }\frac{1-\cos x}{1-\cos x}\)

. . \(\displaystyle \displaystyle\frac{1}{1+\cos x}\cdot\frac{1-\cos x}{1-\cos x} \;=\;\frac{1-\cos x}{1-\cos^2\!x} \;=\;\frac{1-\cos x}{\sin^2\!x}\)

. . \(\displaystyle \displaystyle =\;\frac{1}{\sin^2\!x} - \frac{\cos x}{\sin^2\!x} \;=\;\frac{1}{\sin^2\!x} - \frac{1}{\sin x}\!\cdot\!\frac{\cos x}{\sin x} \;=\;\csc^2\!x - \csc x\cot x\)


\(\displaystyle \displaystyle\text{And we have: }\:\int (\csc^2x - \csc x\cot x)\,dx \;=\;-\cot x + \csc x + C\)

 

[ninguen]

Thanks guys!

Geez calculus is hard. These problems are really tricky. And it doesn't help that the self-teaching resources are undependable and full of errors.

\(\displaystyle \displaystyle \int\frac{1}{1+cos(x)}dx\)

Rewrite as:

\(\displaystyle \displaystyle \int\frac{1}{\underbrace{cos^{2}(x/2)+sin^{2}(x/2)}_{\text{1}}+\underbrace{cos^{2}(x/2)-sin^{2}(x/2)}_{\text{cos(x)}}}dx\)

BTW. I've never seen that trig identity for cos(x). Where did that come from?