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  • integrate x plus sinx by 1 plus cos x dx equals

\int \frac{x+\sin x}{1+\cos x} d x is equal to

\\A. \log |1+\cos x|+c\\ B. \log |\mathrm{x}+\sin \mathrm{x}|+\mathrm{C}\\ C. x-\tan \frac{x}{2}+C\\ D. x \cdot \tan \frac{x}{2}+C\\

Answers (1)

D) \ x \cdot \tan \frac{x}{2}+C\\

Given: \int \frac{x+\sin x}{1+\cos x} d x

As we know

\sin 2 \mathrm{x}=\frac{2 \tan \mathrm{x}}{1+\tan ^{2} \mathrm{x}}, 1+\tan ^{2} \mathrm{x}=\sec ^{2} \mathrm{x} \text { and } \cos 2 \mathrm{x}=\frac{1+\tan ^{2} \mathrm{x}}{1-\tan ^{2} \mathrm{x}}

\Rightarrow \frac{x+\sin x}{1+\cos x}=\frac{x+\frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}}{1+\frac{1+\tan ^{2}\left(\frac{x}{2}\right)}{1-\tan ^{2}\left(\frac{x}{2}\right)}}

\\ \Rightarrow=\frac{x+x \tan ^{2}\left(\frac{x}{2}\right)+2 \tan \left(\frac{x}{2}\right)}{2} \\ \Rightarrow \int \frac{x+\sin x}{1+\cos x} d x=\int \frac{x+x \tan ^{2}\left(\frac{x}{2}\right)+2 \tan \left(\frac{x}{2}\right)}{2} d x \\ \Rightarrow \quad \operatorname{let} \frac{x}{2}=t \\ \Rightarrow \frac{d x}{2}=d t

 

\begin{aligned} &\Rightarrow \int\left(2 t+2 t \tan ^{2} t+2 \tan t\right) d t=2 \int\left(t+t \tan ^{2} t+\tan t\right) d t \\ &\Rightarrow 2 \int tdt+2 \int t\left(\sec ^{2}t-1\right) dt+2 \int \tan t dt \\ &\Rightarrow 2 \int tdt+2 \int t \sec ^{2} t dt-2 \int tdt+2 \int \tan t dt \\ &\Rightarrow 2 \int t \sec ^{2} t dt+2 \int \tan t dt \ldots (1) \end{aligned}

\\\text{Applying Integration by parts on } \int \mathrm{t} \sec ^{2} \mathrm{t} dt \\ \Rightarrow \int t \sec ^{2} t d t=t \int \sec ^{2} t d t-\int\left(\frac{d}{d t} t\right)\left(\int \sec ^{2} t d t\right) d t

\begin{aligned} &\Rightarrow \int t \sec^2 t dt = t \tan t - \int \tan t dt \ldots (2) \end{aligned}

\\ \begin{aligned} &\Rightarrow \text { Put }(2) \text { in }(1)\\ &\Rightarrow 2\left(\mathrm{t} \text { tant }-\int \tan t \mathrm{dt}\right)+2 \int \operatorname{tant} \mathrm{dt}=2 \mathrm{t} \operatorname{tant}+\mathrm{c}=\operatorname{xtan}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}\\ &\Rightarrow \int \frac{x+\sin x}{1+\cos x} d x=\operatorname{xtan}\left(\frac{x}{2}\right)+c \end{aligned}

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