Test Index

UPSC NDA Math Model Paper 9

© examsnet.com

Question : 27 of 120

Marks: +1, -0

The value of

\int\limits_{0}^{1} \sqrt{1+\sin\frac{x}{2}} \, dx

is

0
2
4
None of these

Solution:

Let,

I=\int\limits_{0}^{1} \sqrt{1+\sin\frac{x}{2}} \, dx

=\int\limits_{0}^{1} \sqrt{\sin^{2}\frac{x}{4}+\cos^{2}\frac{x}{4}+\sin\frac{x}{2}} \, dx

\left(\sin^{2}\frac{x}{4}+\cos^{2}\frac{x}{4}=1\right)

=\int\limits_{0}^{1} \sqrt{\sin^{2}\frac{x}{4}+\cos^{2}\frac{x}{4}+2\sin\frac{x}{4}\cos\frac{x}{4}} \, dx

\left(\sin\frac{x}{2}=2\sin\frac{x}{4}\cos\frac{x}{4}\right)

\Rightarrow I=\int\limits_{0}^{1} \sqrt{\left(\sin\frac{x}{4}+\cos\frac{x}{4}\right)^{2}} \, dx

\Rightarrow I=\int\limits_{0}^{1} \left|\sin\frac{x}{4}+\cos\frac{x}{4}\right| \, dx

Let

\frac{x}{4}=t

Differentiating with respect to

x

, we get

\Rightarrow \frac{dx}{4}=dt

\Rightarrow dx=4 \, dt

\therefore I=4 \int\limits_{0}^{1} \left|\sin t+\cos t\right| \, dt

\Rightarrow I=4[-\cos t+\sin t]_{0}^{1}

=4\left[-\cos\frac{x}{4}+\sin\frac{x}{4}\right]_{0}^{1}

=4\left[\left(-\cos\frac{1}{4}+\sin\frac{1}{4}\right)-(-\cos 0+\sin 0)\right]

=4\left[1-\cos\frac{1}{4}+\sin\frac{1}{4}\right]

Hence, option (4) is correct.

© examsnet.com

Go to Question:

Prev Question

Next Question