Indefinite Integrals
Indefinite Integrals
Evaluation of Integrals \(\displaystyle (a):: \int\frac{1}{\left(2+\cos x\right)^2}dx\) and \(\displaystyle (b):: \int\frac{1}{\left(2+\cos x\right)^3}dx\)
- Tolaso J Kos
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Re: Indefinite Integrals
Straight forward calculations but very tedious:
a.$$\begin{aligned}
\int \frac{dx}{\left ( 2+\cos x \right )^2} &\overset{u=\tan \frac{x}{2}}{=\! =\! =\! =\! =\!}\,2\int \frac{du}{\left ( u^2+1 \right )\left ( \frac{1-u^2}{u^2+1}+2 \right )^2} \\
&= \int \frac{2\left ( u^2+1 \right )}{u^4+6u^2+9}\,du=\int \frac{2\left ( u^2+1 \right )}{\left ( u^2+3 \right )^2}\,du\\
&= 2\int \left ( \frac{1}{u^2+3}-\frac{2}{\left ( u^2+3 \right )^2} \right )\,du\\
&= \frac{2}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}}-4\int \frac{du}{\left ( u^2+3 \right )^2}\\
&\overset{u=\sqrt{3}\tan p}{=\! =\! =\! =\! =\! =\!} \frac{2}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}}-\frac{4}{3\sqrt{3}}\int \cos^2 p\,dp\\
&= \frac{2}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}}-\frac{4}{3\sqrt{3}}\left ( \frac{1}{2}p+\sin p\cos p \right )\\
&= \frac{2}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}}-\frac{4}{3\sqrt{3}}\left ( \tan^{-1}\frac{u}{\sqrt{3}}+\sin \tan^{-1}\frac{u}{\sqrt{3}}\cos \tan^{-1}\frac{u}{\sqrt{3}} \right )
\end{aligned}$$ Substitute back for \(x \) to get: $$\boxed{ \displaystyle \int \frac{dx}{\left ( 2+\cos x \right )^3}=\frac{1}{9}\left ( 4\sqrt{3}\tan^{-1}\frac{\tan \frac{x}{2}}{\sqrt{3}}-3\frac{\sin x}{\cos x+2} \right )+c, \;\; c \in \mathbb{R} }$$
b.Same technic applied here... same comment about the calculations.
Dealing with them with perseverance , patience , and caution one gets: $$\boxed{\displaystyle \int \frac{dx}{\left ( 2+\cos x \right )^3}=\frac{1}{\sqrt{3}}\tan^{-1}\frac{\tan \frac{x}{2}}{\sqrt{3}}-\frac{\sin x \left ( 2\cos x+5 \right )}{6\left ( \cos x+2 \right )^2}+c, \;\; c \in \mathbb{R}}$$
Jacks, I hope this suits your fancy.
a.$$\begin{aligned}
\int \frac{dx}{\left ( 2+\cos x \right )^2} &\overset{u=\tan \frac{x}{2}}{=\! =\! =\! =\! =\!}\,2\int \frac{du}{\left ( u^2+1 \right )\left ( \frac{1-u^2}{u^2+1}+2 \right )^2} \\
&= \int \frac{2\left ( u^2+1 \right )}{u^4+6u^2+9}\,du=\int \frac{2\left ( u^2+1 \right )}{\left ( u^2+3 \right )^2}\,du\\
&= 2\int \left ( \frac{1}{u^2+3}-\frac{2}{\left ( u^2+3 \right )^2} \right )\,du\\
&= \frac{2}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}}-4\int \frac{du}{\left ( u^2+3 \right )^2}\\
&\overset{u=\sqrt{3}\tan p}{=\! =\! =\! =\! =\! =\!} \frac{2}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}}-\frac{4}{3\sqrt{3}}\int \cos^2 p\,dp\\
&= \frac{2}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}}-\frac{4}{3\sqrt{3}}\left ( \frac{1}{2}p+\sin p\cos p \right )\\
&= \frac{2}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}}-\frac{4}{3\sqrt{3}}\left ( \tan^{-1}\frac{u}{\sqrt{3}}+\sin \tan^{-1}\frac{u}{\sqrt{3}}\cos \tan^{-1}\frac{u}{\sqrt{3}} \right )
\end{aligned}$$ Substitute back for \(x \) to get: $$\boxed{ \displaystyle \int \frac{dx}{\left ( 2+\cos x \right )^3}=\frac{1}{9}\left ( 4\sqrt{3}\tan^{-1}\frac{\tan \frac{x}{2}}{\sqrt{3}}-3\frac{\sin x}{\cos x+2} \right )+c, \;\; c \in \mathbb{R} }$$
b.Same technic applied here... same comment about the calculations.
Dealing with them with perseverance , patience , and caution one gets: $$\boxed{\displaystyle \int \frac{dx}{\left ( 2+\cos x \right )^3}=\frac{1}{\sqrt{3}}\tan^{-1}\frac{\tan \frac{x}{2}}{\sqrt{3}}-\frac{\sin x \left ( 2\cos x+5 \right )}{6\left ( \cos x+2 \right )^2}+c, \;\; c \in \mathbb{R}}$$
Jacks, I hope this suits your fancy.
Imagination is much more important than knowledge.
Re: Indefinite Integrals
Thanks Apostolos J. Kos for Nice explanation.
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