Definite IntegrationmediumFree

∫(|cosx|+|sinx|) dx from 0 to nπ+t | JEE

JEE Maths question with a full step-by-step solution.

Question

The value of

\displaystyle\int_{0}^{n\pi+t}\big(|\cos x|+|\sin x|\big)\,dx

, where

0<t<\dfrac{\pi}{2}

, is

n

2n+\sin t+\cot t

\cos t

4n+\sin t-\cos t+1

correct

Solution

Step 1: The integrand

|\cos x|+|\sin x|

has period

\dfrac{\pi}{2}

. Split the interval at

x=n\pi

I=\int_{0}^{n\pi}\big(|\cos x|+|\sin x|\big)\,dx+\int_{n\pi}^{n\pi+t}\big(|\cos x|+|\sin x|\big)\,dx.

Step 2: Evaluate the bulk. The length

n\pi

contains

2n

periods of length

\dfrac{\pi}{2}

, and over one period

\int_{0}^{\pi/2}\big(\cos x+\sin x\big)\,dx=\big[\sin x-\cos x\big]_{0}^{\pi/2}=(1-0)-(0-1)=2.

Hence the bulk equals

2n\cdot 2=4n.

Step 3: Evaluate the tail. For

0<t<\dfrac{\pi}{2}

, on

[n\pi,\,n\pi+t]

both

|\cos x|

and

|\sin x|

reduce to

\cos

and

\sin

of the shifted variable, so shifting by

n\pi

\int_{n\pi}^{n\pi+t}\big(|\cos x|+|\sin x|\big)\,dx=\int_{0}^{t}\big(\cos x+\sin x\big)\,dx=\big[\sin x-\cos x\big]_{0}^{t}=\sin t-\cos t+1.

Step 4: Add:

I=4n+\sin t-\cos t+1.

Correct answer: (4)

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