Integrate the following functions with respect to $x$: (i) $\sqrt{9-(2x+5)^2}$ (ii) $\sqrt{81+(2x+1)^2}$ (iii) $\sqrt{(x+1)^2-4}$

Integrate the following functions with respect to x: (i) √(9-(2x+5)²) (ii) √(81+(2x+1)²) (iii) √((x+1)²-4)

Exercise 11.12 · Q2

x

(i)

\sqrt{9-(2x+5)^2}

(ii)

\sqrt{81+(2x+1)^2}

(iii)

\sqrt{(x+1)^2-4}

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Each radicand is already a Type IV shape $a^2-u^2$ / $a^2+u^2$ / $u^2-a^2$ in a linear $u$ ; substitute $u$ , apply Result 11.3, then rescale by the Jacobian $\tfrac12$ .

(i) Let $I=\displaystyle\int\sqrt{9-(2x+5)^2}\,dx$ .

Step 1. Put $u=2x+5$ , $du=2\,dx$ : $I=\dfrac12\displaystyle\int\sqrt{9-u^2}\,du$ .

Step 2. $\displaystyle\int\sqrt{a^2-u^2}\,du=\dfrac u2\sqrt{a^2-u^2}+\dfrac{a^2}2\sin^{-1}(u/a)+c$ with $a=3$ : $I=\dfrac12\left[\dfrac u2\sqrt{9-u^2}+\dfrac92\sin^{-1}\!\left(\dfrac u3\right)\right]+c=\dfrac u4\sqrt{9-u^2}+\dfrac94\sin^{-1}\!\left(\dfrac u3\right)+c$ .

Step 3. Back-substitute $u=2x+5$ : $I=\dfrac{2x+5}4\sqrt{9-(2x+5)^2}+\dfrac94\sin^{-1}\!\left(\dfrac{2x+5}3\right)+c$ .

(ii) Let $I=\displaystyle\int\sqrt{81+(2x+1)^2}\,dx$ .

Step 1. Put $u=2x+1$ , $du=2\,dx$ : $I=\dfrac12\displaystyle\int\sqrt{81+u^2}\,du$ .

Step 2. With $a=9$ : $I=\dfrac12\left[\dfrac u2\sqrt{81+u^2}+\dfrac{81}2\log\left|u+\sqrt{81+u^2}\right|\right]+c=\dfrac u4\sqrt{81+u^2}+\dfrac{81}4\log\left|u+\sqrt{81+u^2}\right|+c$ .

Step 3. Back-substitute $u=2x+1$ : $I=\dfrac{2x+1}4\sqrt{81+(2x+1)^2}+\dfrac{81}4\log\left|(2x+1)+\sqrt{81+(2x+1)^2}\right|+c$ .

(iii) Let $I=\displaystyle\int\sqrt{(x+1)^2-4}\,dx$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set ANNUAL1 markMCQ
$\displaystyle\int \sqrt{\dfrac{1-x}{1+x}}\, dx$
(a) $\sqrt{1-x^2} + \sin^{-1}x + c$
(b) $\sin^{-1}x - \sqrt{1-x^2} + c$
(c) $\log\left|x + \sqrt{1-x^2}\right| - \sqrt{1-x^2} + c$
(d) $\sqrt{1-x^2} + \log\left|x + \sqrt{1-x^2}\right| + c$
›Reveal solutionSolution

Rationalise the integrand by multiplying by $\sqrt{1-x}/\sqrt{1-x}$ , then split and integrate term by term.

Multiply numerator and denominator inside the square root by $(1-x)$ :

$\sqrt{\dfrac{1-x}{1+x}}=\sqrt{\dfrac{(1-x)^2}{(1+x)(1-x)}}=\dfrac{1-x}{\sqrt{1-x^2}}=\dfrac{1}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}.$

Now integrate each piece: …

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CBSE 2019Set ANNUAL1 markMCQ
$\int \dfrac{\sec x}{\sqrt{\cos 2x}}\,dx$
(a) $\tan^{-1}(\cos x)+c$
(b) $\sin^{-1}(\tan x)+c$
(c) $\tan^{-1}(\sin x)+c$
(d) $2\sin^{-1}(\tan x)+c$
›Reveal solutionSolution

Writing $\cos 2x = \cos^2x(1-\tan^2x)$ converts the integrand into $\sec^2x/\sqrt{1-\tan^2x}$ , and substituting $t=\tan x$ gives a standard $\int dt/\sqrt{1-t^2}$ integral.

$\dfrac{\sec x}{\sqrt{\cos 2x}} = \dfrac{1}{\cos x\sqrt{\cos 2x}}$ .

Use $\cos 2x = \cos^2x - \sin^2x = \cos^2x(1-\tan^2x)$ , so $\sqrt{\cos 2x} = \cos x\sqrt{1-\tan^2x}$ (taking $\cos x>0$ ).

Then $\dfrac{1}{\cos x\cdot\cos x\sqrt{1-\tan^2x}} = \dfrac{\sec^2x}{\sqrt{1-\tan^2x}}$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{e^x}{e^{2x}+1}\,dx$
(a) $\log(e^x+1)+c$
(b) $\log(e^{2x}+1)+c$
(c) $\tan^{-1}(e^{2x})+c$
(d) $\tan^{-1}(e^x)+c$
›Reveal solutionSolution

The substitution $u=e^x$ reduces the integral to $\int\frac{du}{u^2+1} = \tan^{-1}u+c$ , giving $\tan^{-1}(e^x)+c$ .

Let $u = e^x$ , so $du = e^x\,dx$ .

…

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