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Indefinite Integrals question

2018 · 15 Apr · Shift 2 · Q38
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Indefinite Integrals question

2018 · 15 Apr · Shift 2 · Q38

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫2x+57−6x−x2  dx=A7−6x−x2+Bsin⁡−1(x+34)+C\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx = A\sqrt {7 - 6x - {x^2}} + B{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right) + C∫7−6x−x2​2x+5​dx=A7−6x−x2​+Bsin−1(4x+3​)+C (where C is a constant of integration), then the ordered pair (A, B) is equal to :
  1. A
    (2, 1)
  2. B
    (−-− 2, −-− 1)
  3. C
    (−-− 2, 1)
  4. D
    (2, −-− 1)
View written solutionFree

Correct answer: B

  1. Given integral

We need to evaluate I=∫2x+57−6x−x2 dxI=\int \frac{2x+5}{\sqrt{7-6x-x^2}}\,dxI=∫7−6x−x2​2x+5​dx and compare it with A7−6x−x2+Bsin⁡−1 ⁣(x+34)+C.A\sqrt{7-6x-x^2}+B\sin^{-1}\!\left(\frac{x+3}{4}\right)+C.A7−6x−x2​+Bsin−1(4x+3​)+C.

  1. Rewrite the quadratic inside the square root

Complete the square: 7−6x−x2=−(x2+6x−7).7-6x-x^2=-(x^2+6x-7).7−6x−x2=−(x2+6x−7). Now, x2+6x−7=(x+3)2−16.x^2+6x-7=(x+3)^2-16.x2+6x−7=(x+3)2−16. So, 7−6x−x2=16−(x+3)2.7-6x-x^2=16-(x+3)^2.7−6x−x2=16−(x+3)2.

Hence, I=∫2x+516−(x+3)2 dx.I=\int \frac{2x+5}{\sqrt{16-(x+3)^2}}\,dx.I=∫16−(x+3)2​2x+5​dx.

  1. Split the numerator cleverly

Notice that 2x+5=(2x+6)−1=2(x+3)−1.2x+5=(2x+6)-1=2(x+3)-1.2x+5=(2x+6)−1=2(x+3)−1. Thus, I=∫2(x+3)16−(x+3)2 dx−∫116−(x+3)2 dx.I=\int \frac{2(x+3)}{\sqrt{16-(x+3)^2}}\,dx-\int \frac{1}{\sqrt{16-(x+3)^2}}\,dx.I=∫16−(x+3)2​2(x+3)​dx−∫16−(x+3)2​1​dx.

So let I=I1−I2.I=I_1-I_2.I=I1​−I2​.

  1. Evaluate I1I_1I1​

For I1=∫2(x+3)16−(x+3)2 dx,I_1=\int \frac{2(x+3)}{\sqrt{16-(x+3)^2}}\,dx,I1​=∫16−(x+3)2​2(x+3)​dx, let u=16−(x+3)2  ⟹  du=−2(x+3) dx.u=16-(x+3)^2 \implies du=-2(x+3)\,dx.u=16−(x+3)2⟹du=−2(x+3)dx. Therefore, I1=−∫duu=−2u=−216−(x+3)2.I_1=-\int \frac{du}{\sqrt{u}}=-2\sqrt{u}=-2\sqrt{16-(x+3)^2}.I1​=−∫u​du​=−2u​=−216−(x+3)2​. Since 16−(x+3)2=7−6x−x216-(x+3)^2=7-6x-x^216−(x+3)2=7−6x−x2, I1=−27−6x−x2.I_1=-2\sqrt{7-6x-x^2}.I1​=−27−6x−x2​.

  1. Evaluate I2I_2I2​

We use the standard result ∫dxa2−u2=sin⁡−1(ua)+C\int \frac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C∫a2−u2​dx​=sin−1(au​)+C when du=dxdu=dxdu=dx.

Here,

=\sin^{-1}\left(\frac{x+3}{4}\right).$$ Therefore, $$I=I_1-I_2=-2\sqrt{7-6x-x^2}-\sin^{-1}\left(\frac{x+3}{4}\right)+C.$$ --- 6. **Compare with the required form** Given $$I=A\sqrt{7-6x-x^2}+B\sin^{-1}\left(\frac{x+3}{4}\right)+C,$$ we get $$A=-2,\qquad B=-1.$$ Thus, $$\boxed{(A,B)=(-2,-1)}.$$ --- 7. **Check with options** Option **B** is $$(-2,-1),$$ which matches our result. --- 8. **Comparison with stored correct answer** Stored correct answer: **B** Our derived answer also gives **B**, so they agree.
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