Class 12 Maths, Worked in Full Past papersGet updates
CBSE Sample Paper 2022-23 · Section C · 3 marks

Q26: Find ∫dx3−2x−x2\int\frac{dx}{\sqrt{3 - 2x - x^2}}

Sample 2022-23Ch 7 · Integrals
The question

Find: ∫dx3−2x−x2\displaystyle\int\frac{dx}{\sqrt{3 - 2x - x^2}}

Exam answer

Let I=∫dx3−2x−x2I = \int\frac{dx}{\sqrt{3 - 2x - x^2}}⁠.

3−2x−x2=−(x2+2x−3)=−[(x+1)2−4]=4−(x+1)2\begin{aligned} 3 - 2x - x^2 &= -(x^2 + 2x - 3) \\ &= -\left[(x + 1)^2 - 4\right] \\ &= 4 - (x + 1)^2 \end{aligned}
Why?

A quadratic under a square root is handled by completing the square. Take out the minus sign first, so that x2x^2 has coefficient +1+1⁠.

3−2x−x2=−x2−2x+3=−(x2+2x−3)\begin{aligned} 3 - 2x - x^2 &= -x^2 - 2x + 3 \\ &= -(x^2 + 2x - 3) \end{aligned}

Half the coefficient of xx is 1, so add and subtract 121^2⁠:

x2+2x−3=(x2+2x+1)−1−3=(x+1)2−4\begin{aligned} &x^2 + 2x - 3 \\ &= (x^2 + 2x + 1) - 1 - 3 \\ &= (x + 1)^2 - 4 \end{aligned}

Then put the minus sign back in:

−[(x+1)2−4]=−(x+1)2+4=4−(x+1)2\begin{aligned} &-\left[(x + 1)^2 - 4\right] \\ &= -(x + 1)^2 + 4 \\ &= 4 - (x + 1)^2 \end{aligned}
I=∫dx−(x2+2x−3)=∫dx22−(x+1)2\begin{aligned} I &= \int\frac{dx}{\sqrt{-(x^2 + 2x - 3)}} \\ &= \int\frac{dx}{\sqrt{2^2 - (x + 1)^2}} \end{aligned}
Why?

Writing 4 as 222^2 shows the shape of a standard integral,

∫dta2−t2\int\frac{dt}{\sqrt{a^2 - t^2}}

with a=2a = 2 and t=x+1t = x + 1⁠.

Put x+1=tx + 1 = t⁠. Then dx=dtdx = dt⁠.

I=∫dt22−t2=sin⁡−1(t2)+C\begin{aligned} I &= \int\frac{dt}{\sqrt{2^2 - t^2}} \\ &= \sin^{-1}\left(\frac{t}{2}\right) + C \end{aligned}
Why?

Differentiating t=x+1t = x + 1 gives dtdx=1\frac{dt}{dx} = 1⁠, so dx=dtdx = dt and the integral keeps the same shape in tt⁠.

[∫dxa2−x2=sin⁡−1(xa)+C]\left[\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C\right]

Why?

This is the standard result used, here with a=2a = 2 and tt in place of xx⁠. Finally put back t=x+1t = x + 1⁠.

Watch out: a2−x2a^2 - x^2 under the root gives sin⁡−1\sin^{-1}⁠. The form x2−a2x^2 - a^2 gives a log instead.

∴ I=sin⁡−1(x+12)+C\therefore\ I = \sin^{-1}\left(\dfrac{x + 1}{2}\right) + C

Where this comes from

Chapter 7, Integrals: ∫ dx/√(a² − x²) after completing the square.

Question type: Special forms

How to spot it: click to see

Type 4. Special forms: complete the square, then match a card

Spot it: dxdx over a quadratic, or over the square root of a quadratic, with just a number on top.

Here: Under the root, 3−2x−x23 - 2x - x^2 has an xx term, and only 1 is on top. Take out the minus sign and complete the square to get 22−(x+1)22^2 - (x + 1)^2⁠, then put x+1=tx + 1 = t and use the sin⁡−1\sin^{-1} card with a=2a = 2⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Integrals in past papers: question 38 of 46 · see them all