dx/1- x- x is equal to

Correct Answer is : log⁡(1−cot⁡x2)+c≤ft(1-/2)+clog(1−cot2x)+c

Correct Answer is : log⁡(1−cot⁡x2)+c≤ft(1-/2)+clog(1−cot2x)+c

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EAMCET Engineering 2002 Solved Paper

Section: Mathematics

Question : 151 of 200

Marks: +1, -0

\int \frac{dx}{1-\cos x- \sin x}

Solution:

We have,

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

= \int \frac{dx}{1- \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}} - \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}}

= \int \frac{\sec^2\frac{x}{2} dx}{1+\tan^2\frac{x}{2}-1+\tan^2\frac{x}{2}-2\tan\frac{x}{2}}

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

Let

\tan\frac{x}{2}=z

\Rightarrow \frac{1}{2} \sec^2\frac{x}{2} dx = dz

\Rightarrow \sec^2\frac{x}{2} dx = 2 dz

\therefore I = \frac{1}{2} \int \frac{2 dz}{z(z-1)} = \int \left( \frac{1}{z-1} - \frac{1}{z} \right) dz

= \log(z-1)-\log z+C

= \log\left(\frac{z-1}{z}\right)+C = \log\left[ \frac{\tan\frac{x}{2}-1}{\tan\frac{x}{2}} \right]+C

= \log[1-\cot\frac{x}{2}]+C

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