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Indefinite Integrals question

2020 · 9 Jan · Shift 2 · Q26
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Indefinite Integrals question

2020 · 9 Jan · Shift 2 · Q26

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫dθcos⁡2θ(tan⁡2θ+sec⁡2θ)=λtan⁡θ+2log⁡e∣f(θ)∣+C\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C∫cos2θ(tan2θ+sec2θ)dθ​=λtanθ+2loge​∣f(θ)∣+C where C is a constant of integration, then the ordered pair (λ\lambdaλ, ƒ(θ\thetaθ)) is equal to :
  1. A
    (–1, 1 – tan θ\thetaθ)
  2. B
    (1, 1 + tan θ\thetaθ)
  3. C
    (–1, 1 + tan θ\thetaθ)
  4. D
    (1, 1 – tan θ\thetaθ)
View written solutionFree

Correct answer: C

  1. We need to evaluate
∫dθcos⁡2θ (tan⁡2θ+sec⁡2θ).\int \frac{d\theta}{\cos^2\theta\, (\tan 2\theta+\sec 2\theta)}.∫cos2θ(tan2θ+sec2θ)dθ​.
  1. Use the identity tan⁡x+sec⁡x=1+sin⁡xcos⁡x.\tan x+\sec x=\frac{1+\sin x}{\cos x}.tanx+secx=cosx1+sinx​. So, tan⁡2θ+sec⁡2θ=1+sin⁡2θcos⁡2θ.\tan 2\theta+\sec 2\theta=\frac{1+\sin 2\theta}{\cos 2\theta}.tan2θ+sec2θ=cos2θ1+sin2θ​. Hence

Therefore,

  1. Put t=tan⁡θ  ⟹  dt=sec⁡2θ dθ.t=\tan\theta \implies dt=\sec^2\theta\, d\theta.t=tanθ⟹dt=sec2θdθ. Also, sin⁡2θ=2t1+t2,cos⁡2θ=1−t21+t2.\sin 2\theta=\frac{2t}{1+t^2}, \qquad \cos 2\theta=\frac{1-t^2}{1+t^2}.sin2θ=1+t22t​,cos2θ=1+t21−t2​. Thus

Simplify the denominator:

So,

Now factor: 1−t2=(1−t)(1+t).1-t^2=(1-t)(1+t).1−t2=(1−t)(1+t). Hence

So,

  1. Rewrite the integrand: 1−t1+t=(1+t)−2t1+t=1−2t1+t.\frac{1-t}{1+t}=\frac{(1+t)-2t}{1+t}=1-\frac{2t}{1+t}.1+t1−t​=1+t(1+t)−2t​=1−1+t2t​. Also, t1+t=1−11+t.\frac{t}{1+t}=1-\frac{1}{1+t}.1+tt​=1−1+t1​. Therefore,

Thus,

=−t+2ln⁡∣1+t∣+C.=-t+2\ln|1+t|+C.=−t+2ln∣1+t∣+C.

Substituting back t=tan⁡θt=\tan\thetat=tanθ,

  1. Compare with the given form I=λtan⁡θ+2log⁡e∣f(θ)∣+C.I=\lambda \tan\theta+2\log_e|f(\theta)|+C.I=λtanθ+2loge​∣f(θ)∣+C. So, λ=−1,f(θ)=1+tan⁡θ.\lambda=-1, \qquad f(\theta)=1+\tan\theta.λ=−1,f(θ)=1+tanθ.

  2. Therefore the ordered pair is (−1,1+tan⁡θ).(-1,1+\tan\theta).(−1,1+tanθ). This is option C.

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