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The question is to evaluate $$\int_0^{\pi /2} \frac{ \log (1+\cos a \cos x)}{\cos x} dx$$

I tried using leibnitz rule

$$F'(a)=\int_0^{\pi /2} \frac{ -\sin a}{(1+\cos a \cos x)}dx$$ Now I used the substitution $\tan(x/2)=t$ to get $$-2 \sin a \int_0^1 \frac{ dt}{1+t^2 +\cos a (1-t^2)} $$ which can be rewritten as

$$-2\frac{\sin a} {1- \cos a}\int_0^{1} \frac{ dt}{t^2 +\frac{1+ \cos a}{1-\cos a}} $$ which evaluates to $-a$.i am not sure where I went wrong.Any ideas?

Guy Fsone
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user471651
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3 Answers3

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Hint: Put $b= \sqrt{\frac{1+ \cos a}{1-\cos a} }$ Then,

$$\int_0^{1} \frac{ dt}{t^2 +\frac{1+ \cos a}{1-\cos a}} =\int_0^{1} \frac{ dt}{t^2 +b^2} = \frac1b\arctan{\frac{t}{b}}\bigg|_{t=0}^{t=1}= \frac1b\arctan{\frac{1}{b}} $$

Therefore, $$ F'(a) =-2\frac{\sin a} {\sqrt{(1-\cos a)(1+ \cos a)}} \arctan{\left(\sqrt{\frac{1- \cos a}{1+\cos a} }\right)} \\= -2\frac{\sin a} {|\sin a|} \arctan{\left(\sqrt{\frac{1- \cos a}{1+\cos a} }\right)}$$

That is $$ F'(a)= \color{blue}{ -2\frac{\sin a} {|\sin a|} \arctan{\left(\sqrt{\frac{1- \cos a}{1+\cos a} }\right)}}$$

Guy Fsone
  • 23,903
2

Well, as you did:

$$\mathscr{I}:=\int\frac{1}{1+\text{n}\cdot\cos\left(x\right)}\space\text{d}x\tag1$$

Substitute $\text{u}:=\tan\left(\frac{x}{2}\right)$:

$$\mathscr{I}=\frac{2}{1+\text{n}}\cdot\int\frac{1}{1+\frac{1-\text{n}}{1+\text{n}}\cdot\text{u}^2}\space\text{d}\text{u}\tag2$$

Substitute $\text{s}:=\text{u}\cdot\sqrt{\frac{1-\text{n}}{1+\text{n}}}$:

$$\mathscr{I}=\frac{2}{\sqrt{1-\text{n}}\cdot\sqrt{1+\text{n}}}\cdot\int\frac{1}{1+\text{s}^2}\space\text{d}\text{s}=\frac{2\cdot\arctan\left(\text{s}\right)}{\sqrt{1-\text{n}}\cdot\sqrt{1+\text{n}}}+\text{C}=$$ $$\frac{2\cdot\arctan\left(\tan\left(\frac{x}{2}\right)\cdot\sqrt{\frac{1-\text{n}}{1+\text{n}}}\right)}{\sqrt{1-\text{n}}\cdot\sqrt{1+\text{n}}}+\text{C}\tag3$$

So, for the definite integral:

$$\int_0^\frac{\pi}{2}\frac{1}{1+\text{n}\cdot\cos\left(x\right)}\space\text{d}x=\frac{2\cdot\arctan\left(\sqrt{\frac{2}{\text{n}^2}-1}\right)}{\sqrt{1-\text{n}^2}}\tag4$$

Jan Eerland
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Using Feynman’s Technique Integration by differentiating w.r.t. $a$, $$I(a)=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos a \cos x)}{\cos x} d x,$$

we obtain $$ I^{\prime}(a)=\int_{0}^{\frac{\pi}{2}} \frac{-\sin a}{1+\cos a \cos x} d x $$ Letting $t=\tan \dfrac{x}{2}$ yields $$ \begin{aligned} I^{\prime}(a) &=-2 \sin a \int_{0}^{1} \frac{d t}{(1+\cos a)+(1-\cos a) t^{2}} \\ &=-2 \sin a \int_{0}^{1} \frac{d t}{2 \cos ^{2} \frac{a}{2}+2 t^{2} \sin ^{2} \frac{a}{2}} \\ &=-2 \cos \frac{a}{2} \int_{0}^{1} \frac{d\left(t \sin \frac{a}{2}\right)}{\left(t \sin \frac{a}{2}\right)^{2}+\cos ^{2} \frac{a}{2}} \\ &=-2\left[\tan ^{-1}\left(t \cdot \tan \frac{a}{2}\right)\right]_{0}^{1} \\ &=-2 \tan ^{-1}\left(\tan \frac{a}{2}\right) \\ &=-a \end{aligned} $$ Now we can get $I(a)$ by integrating $I^{\prime}(y)$ w.r.t $y$ from $\dfrac{\pi}{2} $ to $a$.

$$ I(a)=\int_{\frac{\pi}{2}}^{a} I^{\prime}(y) d y=\int_{\frac{\pi}{2}}^{a}-yd y=\boxed{\frac{\pi^{2}}{8}-\frac{a^{2}}{2}} $$ For example, $$ \int_{0}^{\frac{\pi}{2}} \frac{\ln \left(1+\frac{1}{2} \cos x\right)}{\cos x} d x=\frac{5 \pi^{2}}{72} $$ :|D Wish you enjoy my solution!

Lai
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