Test Index

AP EAMCET Engineering 2013 Solved Paper

Section: Mathematics

© examsnet.com

Question : 154 of 160

Marks: +1, -0

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

+p \log \left|\sec \left(\frac{x}{2}\right)\right|+C

p

is equal to

$-4$
4
2
$-2$

Solution:

Let

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

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

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

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

-\int \tan\left(\frac{x}{2}\right) dx

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

-\int \tan\left(\frac{x}{2}\right) + C

= x \tan\left(\frac{x}{2}\right) - 4 \log \left|\sec\left(\frac{x}{2}\right)\right| + C

but given,

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

= x \tan\left(\frac{x}{2}\right) + p \log \left|\sec\left(\frac{x}{2}\right)\right| + C

On comparing, we get

p = -4

© examsnet.com

Go to Question:

Prev Question

Next Question