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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Let ∫ (2x) ( cos x ) dx = f ( x ) ^ 2 x +C, (where, C is the...

Let∫sin⁡(2x)ln⁡(cosx)dx=f(x)cos⁡2x+C,∫ \sin \left(2x\right)\ln ( cos x ) dx = f ( x ) \cos^{2} x +C, (where, CC is the constant of integration)andf(0)=12.and f\left(0\right) = \dfrac{1}{2} . If f(π3)f\left(\dfrac{\pi}{3}\right) is equal to 1a+lnb,\dfrac{1}{a} + ln b, then the value of a+ba + b is

Correct answer

4

Step-by-step solution

Given integral is
I=∫2sinxcosxln(cosx)dxI = ∫ 2sin xcos xln ( cos x ) dx
Let, cosx=tcos x = t
⇒−sinxdx=dt⇒ - sin xdx = dt
⇒I=−2∫tlntdt⇒ I = - 2∫ tln tdt
Using Integration by parts
I=−2(t22lnt−∫t221tdt)I = - 2\left(\dfrac{{t}^{2}}{2}ln t - ∫ \dfrac{{t}^{2}}{2}\dfrac{1}{t}dt\right)
I=−t2lnt+t22+CI = - {t}^{2}ln t + \dfrac{{t}^{2}}{2} + C
I=cos⁡2x(12−lncosx)+CI = \cos^{2} x\left(\dfrac{1}{2} - ln cos x\right) + C
f(x)=12−ln⁡(cosx)f\left(x\right) = \dfrac{1}{2} - \ln \left(cos x\right)
f(π3)=12−ln⁡(12)f\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2} - \ln \left(\dfrac{1}{2}\right)
=12+ln⁡2= \dfrac{1}{2} + \ln 2
⇒a=b=2⇒ a = b = 2
∴a+b=4∴ a + b = 4
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