Exercise 7.7 Practice
Integrals of some more types – NCERT Solutions
Q1
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Integrate the function: $\sqrt{4-x^2}$
Formula: $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$
Here, $\sqrt{4-x^2} = \sqrt{2^2-x^2}$. So, $a=2$.
Substituting $a=2$ in the formula:
$\boxed{\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2} + C}$
Q2
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Integrate the function: $\sqrt{1-4x^2}$
Let $I = \int \sqrt{1-4x^2} dx = \int \sqrt{1-(2x)^2} dx$.
Put $2x = t \Rightarrow 2dx = dt \Rightarrow dx = \frac{dt}{2}$.
$I = \frac{1}{2} \int \sqrt{1^2-t^2} dt$. Using $\int \sqrt{a^2-x^2} dx$ with $a=1$:
$= \frac{1}{2} \left[ \frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin^{-1}t \right] + C$
Substitute $t=2x$:
$\boxed{\frac{x}{2}\sqrt{1-4x^2} + \frac{1}{4}\sin^{-1}(2x) + C}$
Q3
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Integrate the function: $\sqrt{x^2+4x+6}$
Completing the square: $x^2+4x+6 = x^2 + 4x + 4 + 2 = (x+2)^2 + (\sqrt{2})^2$.
Let $I = \int \sqrt{(x+2)^2 + (\sqrt{2})^2} dx$.
Using $\int \sqrt{x^2+a^2} dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| + C$:
Here $X = x+2$ and $a = \sqrt{2}$.
$\boxed{\frac{x+2}{2}\sqrt{x^2+4x+6} + \log|x+2+\sqrt{x^2+4x+6}| + C}$
Q4
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Integrate the function: $\sqrt{x^2+4x+1}$
Completing the square: $x^2+4x+1 = x^2 + 4x + 4 - 3 = (x+2)^2 - (\sqrt{3})^2$.
Let $I = \int \sqrt{(x+2)^2 - (\sqrt{3})^2} dx$.
Using $\int \sqrt{x^2-a^2} dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| + C$:
Here $X = x+2$ and $a = \sqrt{3}$.
$\boxed{\frac{x+2}{2}\sqrt{x^2+4x+1} - \frac{3}{2}\log|x+2+\sqrt{x^2+4x+1}| + C}$
Q5
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Integrate the function: $\sqrt{1-4x-x^2}$
Completing the square: $1-4x-x^2 = 1-(x^2+4x) = 1-(x^2+4x+4-4) = 1-(x+2)^2+4 = 5-(x+2)^2$.
Let $I = \int \sqrt{(\sqrt{5})^2 - (x+2)^2} dx$.
Using $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$:
Here $a=\sqrt{5}$ and $X=x+2$.
$\boxed{\frac{x+2}{2}\sqrt{1-4x-x^2} + \frac{5}{2}\sin^{-1}(\frac{x+2}{\sqrt{5}}) + C}$
Q6
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Integrate the function: $\sqrt{x^2+4x-5}$
Completing the square: $x^2+4x-5 = x^2+4x+4-9 = (x+2)^2 - 3^2$.
Let $I = \int \sqrt{(x+2)^2 - 3^2} dx$.
Using $\int \sqrt{x^2-a^2} dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| + C$:
Here $X=x+2$ and $a=3$.
$\boxed{\frac{x+2}{2}\sqrt{x^2+4x-5} - \frac{9}{2}\log|x+2+\sqrt{x^2+4x-5}| + C}$
Q7
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Integrate the function: $\sqrt{1+3x-x^2}$
Completing the square: $1+3x-x^2 = 1-(x^2-3x) = 1-(x^2-3x+\frac{9}{4}-\frac{9}{4}) = 1+\frac{9}{4}-(x-\frac{3}{2})^2 = \frac{13}{4}-(x-\frac{3}{2})^2$.
Let $I = \int \sqrt{(\frac{\sqrt{13}}{2})^2 - (x-\frac{3}{2})^2} dx$.
Using $\int \sqrt{a^2-x^2} dx$: Here $a=\frac{\sqrt{13}}{2}$ and $X=x-\frac{3}{2}$.
$= \frac{x-3/2}{2}\sqrt{1+3x-x^2} + \frac{13/4}{2}\sin^{-1}\left(\frac{x-3/2}{\sqrt{13}/2}\right) + C$
$\boxed{\frac{2x-3}{4}\sqrt{1+3x-x^2} + \frac{13}{8}\sin^{-1}(\frac{2x-3}{\sqrt{13}}) + C}$
Q8
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Integrate the function: $\sqrt{x^2+3x}$
Completing the square: $x^2+3x = x^2+3x+\frac{9}{4}-\frac{9}{4} = (x+\frac{3}{2})^2 - (\frac{3}{2})^2$.
Let $I = \int \sqrt{(x+\frac{3}{2})^2 - (\frac{3}{2})^2} dx$.
Using $\int \sqrt{x^2-a^2} dx$: Here $X=x+\frac{3}{2}$ and $a=\frac{3}{2}$.
$= \frac{x+3/2}{2}\sqrt{x^2+3x} - \frac{9/4}{2}\log|x+\frac{3}{2}+\sqrt{x^2+3x}| + C$
$\boxed{\frac{2x+3}{4}\sqrt{x^2+3x} - \frac{9}{8}\log|x+\frac{3}{2}+\sqrt{x^2+3x}| + C}$
Q9
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Integrate the function: $\sqrt{1+\frac{x^2}{9}}$
Rewrite: $\sqrt{1+\frac{x^2}{9}} = \sqrt{\frac{9+x^2}{9}} = \frac{1}{3}\sqrt{x^2+9}$.
Let $I = \frac{1}{3} \int \sqrt{x^2+3^2} dx$.
Using $\int \sqrt{x^2+a^2} dx$ with $a=3$:
$= \frac{1}{3} \left[ \frac{x}{2}\sqrt{x^2+9} + \frac{9}{2}\log|x+\sqrt{x^2+9}| \right] + C$
$\boxed{\frac{x}{6}\sqrt{x^2+9} + \frac{3}{2}\log|x+\sqrt{x^2+9}| + C}$
Q10
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$\int \sqrt{1+x^2} \, dx$ is equal to:
(A) $\frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log|x+\sqrt{1+x^2}| + C$
(B) $\frac{2}{3}(1+x^2)^{3/2} + C$
(C) $\frac{2}{3}x(1+x^2)^{3/2} + C$
(D) $\frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log|x+\sqrt{1+x^2}| + C$
(A) $\frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log|x+\sqrt{1+x^2}| + C$
(B) $\frac{2}{3}(1+x^2)^{3/2} + C$
(C) $\frac{2}{3}x(1+x^2)^{3/2} + C$
(D) $\frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log|x+\sqrt{1+x^2}| + C$
This is a standard integral of the form $\int \sqrt{x^2+a^2} dx$ with $a=1$.
Formula: $\frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| + C$.
$\boxed{\text{(A) } \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log|x+\sqrt{1+x^2}| + C}$
Q11
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$\int \sqrt{x^2-8x+7} \, dx$ is equal to:
(A) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} + 9\log|x-4+\sqrt{x^2-8x+7}| + C$
(B) $\frac{1}{2}(x+4)\sqrt{x^2-8x+7} + 9\log|x+4+\sqrt{x^2-8x+7}| + C$
(C) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} - \frac{9}{2}\log|x-4+\sqrt{x^2-8x+7}| + C$
(D) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} - \frac{3}{2}\log|x-4+\sqrt{x^2-8x+7}| + C$
(A) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} + 9\log|x-4+\sqrt{x^2-8x+7}| + C$
(B) $\frac{1}{2}(x+4)\sqrt{x^2-8x+7} + 9\log|x+4+\sqrt{x^2-8x+7}| + C$
(C) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} - \frac{9}{2}\log|x-4+\sqrt{x^2-8x+7}| + C$
(D) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} - \frac{3}{2}\log|x-4+\sqrt{x^2-8x+7}| + C$
Completing the square: $x^2-8x+7 = x^2-8x+16-9 = (x-4)^2 - 3^2$.
Let $I = \int \sqrt{(x-4)^2 - 3^2} dx$.
Using $\int \sqrt{x^2-a^2} dx$ with $X=x-4$ and $a=3$:
$\boxed{\text{(D) } \frac{1}{2}(x-4)\sqrt{x^2-8x+7} - \frac{9}{2}\log|x-4+\sqrt{x^2-8x+7}| + C}$