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Indefinite Integrals question

2021 · 24 Feb · Shift 1 · Q23
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  5. /2021 · 24 Feb · Shift 1 · Q23

Indefinite Integrals question

2021 · 24 Feb · Shift 1 · Q23

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫cos⁡x−sin⁡x8−sin⁡2xdx=asin⁡−1(sin⁡x+cos⁡xb)+c\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c∫8−sin2x​cosx−sinx​dx=asin−1(bsinx+cosx​)+c, where c is a constant of integration, then the ordered pair (a, b) is equal to :
  1. A
    (-1, 3)
  2. B
    (1, 3)
  3. C
    (1, -3)
  4. D
    (3, 1)
View written solutionFree

Correct answer: B

  1. Given integral

We need to evaluate

I=∫cos⁡x−sin⁡x8−sin⁡2x dxI=\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}\,dxI=∫8−sin2x​cosx−sinx​dx

and compare it with

asin⁡−1(sin⁡x+cos⁡xb)+c.a\sin^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c.asin−1(bsinx+cosx​)+c.
  1. Use a standard substitution

Notice that

ddx(sin⁡x+cos⁡x)=cos⁡x−sin⁡x.\frac{d}{dx}(\sin x+\cos x)=\cos x-\sin x.dxd​(sinx+cosx)=cosx−sinx.

So let

t=sin⁡x+cos⁡x.t=\sin x+\cos x.t=sinx+cosx.

Then

dt=(cos⁡x−sin⁡x)dx.dt=(\cos x-\sin x)dx.dt=(cosx−sinx)dx.

Also,

(sin⁡x+cos⁡x)2=sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=1+sin⁡2x.(\sin x+\cos x)^2=\sin^2x+\cos^2x+2\sin x\cos x=1+\sin 2x.(sinx+cosx)2=sin2x+cos2x+2sinxcosx=1+sin2x.

Hence,

sin⁡2x=t2−1.\sin 2x=t^2-1.sin2x=t2−1.

Therefore,

8−sin⁡2x=8−(t2−1)=9−t2.8-\sin 2x=8-(t^2-1)=9-t^2.8−sin2x=8−(t2−1)=9−t2.

So the integral becomes

I=∫dt9−t2.I=\int \frac{dt}{\sqrt{9-t^2}}.I=∫9−t2​dt​.
  1. Integrate

Using the standard formula

∫dta2−t2=sin⁡−1(ta)+C,\int \frac{dt}{\sqrt{a^2-t^2}}=\sin^{-1}\left(\frac{t}{a}\right)+C,∫a2−t2​dt​=sin−1(at​)+C,

with a=3a=3a=3, we get

I=sin⁡−1(t3)+C.I=\sin^{-1}\left(\frac{t}{3}\right)+C.I=sin−1(3t​)+C.

Substituting back t=sin⁡x+cos⁡xt=\sin x+\cos xt=sinx+cosx,

I=sin⁡−1(sin⁡x+cos⁡x3)+C.I=\sin^{-1}\left(\frac{\sin x+\cos x}{3}\right)+C.I=sin−1(3sinx+cosx​)+C.

Comparing with

asin⁡−1(sin⁡x+cos⁡xb)+c,a\sin^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c,asin−1(bsinx+cosx​)+c,

we obtain

a=1,b=3.a=1,\qquad b=3.a=1,b=3.
  1. Check options
  • A: (−1,3)(-1,3)(−1,3) ✗
  • B: (1,3)(1,3)(1,3) ✓
  • C: (1,−3)(1,-3)(1,−3) ✗
  • D: (3,1)(3,1)(3,1) ✗

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They match.

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