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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

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Question : 72 of 160

Marks: +1, -0

∫

\frac{x+\sin x}{1+\cos x}

dx =

$\log_e$ (1 + cos x) + x
x $\sin^2 \frac{x}{2}$ + c
tan $\frac{x}{2}$ + c
x tan $\frac{x}{2}$ + c

Solution:

Consider

I \;\; =\int \frac{x+\sin x}{1+\cos x} dx

\;\; =\int \frac{x}{1+\cos x} dx+\int \frac{\sin x}{1+\cos x} dx

Now

I_{1}=\int \frac{x}{1+\cos x} dx = \int \frac{x dx}{1+\left(\frac{1-\tan^2 \frac{x}{2}}{1-\tan^2 \frac{x}{2}}\right)}

=\int \frac{x \sec^{2} \frac{x}{2} dx}{2} \; \frac{x}{2}

\;\; =\int \frac{x}{2} \sec^{2} \frac{x}{2} dx

\;\; =2 \frac{x}{2} \tan \frac{x}{2} - 2 \log \left|\sec \frac{x}{2}\right|+c_{1}

\;\; =x \tan \frac{x}{2} - \log \left|\sec^{2} \frac{x}{2}\right|+c_{1}

Now,

I_{2}=\int \frac{\sin x}{1+\cos x} dx

I_{2}=-\log |1+\cos x|+c_{2}

I_{2}=\log \left| \frac{1}{1+\cos x} \right|+c_{2}

I_{2}=\log \left| \frac{1}{1+\cos^{2} \frac{x}{2}} \right|+c_{2}

I_{2}=\log \left|\sec^{2} \frac{x}{2}\right|+c_{2}

Now

Substitute the value in equation (I).

I=x \tan \frac{x}{2} - \log \left|\sec^{2} \frac{x}{2}\right| + c_{1} + \log \left|\sec^{2} \frac{x}{2}\right| + c_{2}

I=x \tan \frac{x}{2}+c

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