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Indefinite Integrals question

2019 · 9 Jan · Shift 1 · Q39
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  5. /2019 · 9 Jan · Shift 1 · Q39

Indefinite Integrals question

2019 · 9 Jan · Shift 1 · Q39

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
For x2 eee n π\piπ+ 1, n ∈\in∈ N (the set of natural numbers), the integral ∫x2sin⁡(x2−1)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1)dx\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx}∫x2sin(x2−1)+sin2(x2−1)2sin(x2−1)−sin2(x2−1)​​dx is equal to : (where c is a constant of integration)
  1. A
    log⁡e∣12sec⁡2(x2−1)∣+c{\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + cloge​​21​sec2(x2−1)​+c
  2. B
    12log⁡e∣sec⁡(x2−1)∣+c{1 \over 2}{\log _e}\left| {\sec \left( {{x^2} - 1} \right)} \right| + c21​loge​​sec(x2−1)​+c
  3. C
    12log⁡e∣sec⁡2(x2−12)∣+c{1 \over 2}{\log _e}\left| {{{\sec }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \right| + c21​loge​​sec2(2x2−1​)​+c
  4. D
    log⁡e∣sec⁡(x2−12)∣+c{\log _e}\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + cloge​​sec(2x2−1​)​+c
View written solutionFree

Correct answer: D

  1. Given integral

We need to evaluate

I=∫x2sin⁡(x2−1)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1) dx.I=\int x\sqrt{\frac{2\sin(x^2-1)-\sin 2(x^2-1)}{2\sin(x^2-1)+\sin 2(x^2-1)}}\,dx.I=∫x2sin(x2−1)+sin2(x2−1)2sin(x2−1)−sin2(x2−1)​​dx.

Let

t=x2−1  ⟹  dt=2x dx  ⟹  x dx=dt2.t=x^2-1 \implies dt=2x\,dx \implies x\,dx=\frac{dt}{2}.t=x2−1⟹dt=2xdx⟹xdx=2dt​.

So the integral becomes

I=12∫2sin⁡t−sin⁡2t2sin⁡t+sin⁡2t dt.I=\frac12\int \sqrt{\frac{2\sin t-\sin 2t}{2\sin t+\sin 2t}}\,dt.I=21​∫2sint+sin2t2sint−sin2t​​dt.
  1. Simplify the trigonometric expression

Use

sin⁡2t=2sin⁡tcos⁡t.\sin 2t=2\sin t\cos t.sin2t=2sintcost.

Then

2sin⁡t−sin⁡2t=2sin⁡t−2sin⁡tcos⁡t=2sin⁡t(1−cos⁡t),2\sin t-\sin 2t=2\sin t-2\sin t\cos t=2\sin t(1-\cos t),2sint−sin2t=2sint−2sintcost=2sint(1−cost),

and

2sin⁡t+sin⁡2t=2sin⁡t+2sin⁡tcos⁡t=2sin⁡t(1+cos⁡t).2\sin t+\sin 2t=2\sin t+2\sin t\cos t=2\sin t(1+\cos t).2sint+sin2t=2sint+2sintcost=2sint(1+cost).

Hence

2sin⁡t−sin⁡2t2sin⁡t+sin⁡2t=2sin⁡t(1−cos⁡t)2sin⁡t(1+cos⁡t)=1−cos⁡t1+cos⁡t.\frac{2\sin t-\sin 2t}{2\sin t+\sin 2t} =\frac{2\sin t(1-\cos t)}{2\sin t(1+\cos t)} =\frac{1-\cos t}{1+\cos t}.2sint+sin2t2sint−sin2t​=2sint(1+cost)2sint(1−cost)​=1+cost1−cost​.

Now use the identity

1−cos⁡t1+cos⁡t=tan⁡2t2.\frac{1-\cos t}{1+\cos t}=\tan^2\frac t2.1+cost1−cost​=tan22t​.

Therefore,

2sin⁡t−sin⁡2t2sin⁡t+sin⁡2t=tan⁡t2\sqrt{\frac{2\sin t-\sin 2t}{2\sin t+\sin 2t}}=\tan\frac t22sint+sin2t2sint−sin2t​​=tan2t​

under the given domain condition (so the principal square root is consistent).

Thus

I=12∫tan⁡t2 dt.I=\frac12\int \tan\frac t2\,dt.I=21​∫tan2t​dt.
  1. Integrate

Let

u=t2  ⟹  dt=2 dν.u=\frac t2 \implies dt=2\,d\nu.u=2t​⟹dt=2dν.

Then

I=12∫tan⁡t2 dt=12∫tan⁡ν (2 dν)=∫tan⁡ν dν.I=\frac12\int \tan\frac t2\,dt =\frac12\int \tan \nu\,(2\,d\nu) =\int \tan \nu\,d\nu.I=21​∫tan2t​dt=21​∫tanν(2dν)=∫tanνdν.

We know

∫tan⁡ν dν=log⁡e∣sec⁡ν∣+c.\int \tan \nu\,d\nu=\log_e|\sec \nu|+c.∫tanνdν=loge​∣secν∣+c.

So

I=log⁡e∣sec⁡t2∣+c.I=\log_e\left|\sec\frac t2\right|+c.I=loge​​sec2t​​+c.

Substituting back t=x2−1t=x^2-1t=x2−1,

I=log⁡e∣sec⁡(x2−12)∣+c.I=\log_e\left|\sec\left(\frac{x^2-1}{2}\right)\right|+c.I=loge​​sec(2x2−1​)​+c.
  1. Match with options

This is exactly Option D:

log⁡e∣sec⁡(x2−12)∣+c.\log_e\left|\sec\left(\frac{x^2-1}{2}\right)\right|+c.loge​​sec(2x2−1​)​+c.
  1. Verification with stored answer

Stored correct answer: D.

Our derived answer is also D, so they agree.

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