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

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

Question
The value of ∫0nπ+t(∣cos⁡x∣+∣sin⁡x∣) dx\displaystyle\int_{0}^{n\pi+t}\big(|\cos x|+|\sin x|\big)\,dx, where 0<t<π20<t<\dfrac{\pi}{2}, is
Ann
B2n+sin⁡t+cot⁡t2n+\sin t+\cot t
Ccos⁡t\cos t
D4n+sin⁡t−cos⁡t+14n+\sin t-\cos t+1correct
Solution
Step 1: The integrand ∣cos⁡x∣+∣sin⁡x∣|\cos x|+|\sin x| has period π2\dfrac{\pi}{2}. Split the interval at x=nπx=n\pi:
I=∫0nπ(∣cos⁡x∣+∣sin⁡x∣) dx+∫nπnπ+t(∣cos⁡x∣+∣sin⁡x∣) dx.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 nπn\pi contains 2n2n periods of length π2\dfrac{\pi}{2}, and over one period
∫0π/2(cos⁡x+sin⁡x) dx=[sin⁡x−cos⁡x]0π/2=(1−0)−(0−1)=2.\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⋅2=4n.2n\cdot 2=4n. Step 3: Evaluate the tail. For 0<t<π20<t<\dfrac{\pi}{2}, on [nπ, nπ+t][n\pi,\,n\pi+t] both ∣cos⁡x∣|\cos x| and ∣sin⁡x∣|\sin x| reduce to cos⁡\cos and sin⁡\sin of the shifted variable, so shifting by nπn\pi,
∫nπnπ+t(∣cos⁡x∣+∣sin⁡x∣) dx=∫0t(cos⁡x+sin⁡x) dx=[sin⁡x−cos⁡x]0t=sin⁡t−cos⁡t+1.\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.I=4n+\sin t-\cos t+1.
 Correct answer: (4)
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