NCERT Solutions for Exercise 7.7 Class 12 Maths Chapter 7 - Integrals

Question 1: Integrate the functions in Exercises 1 to 9.

$\sqrt{4 - x^2}$

Answer:

Given function $\sqrt{4 - x^2}$ ,

So, let us consider the function to be;

$I = \int \sqrt{4-x^2}dx$

$= \int \sqrt{(2)^2-x^2}dx$

Then it is known that, $= \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

Therefore, $I = \frac{x}{2}\sqrt{4-x^2} +\frac{4}{2}\sin^{-1}{\frac{x}{2}}+C$

$= \frac{x}{2}\sqrt{4-x^2} +2\sin^{-1}{\frac{x}{2}}+C$

Question 2: Integrate the functions in Exercises 1 to 9.

$\sqrt{1 - 4x^2}$

Answer:

Given function to integrate $\sqrt{1 - 4x^2}$

Now we can rewrite as

$= \int \sqrt{1 - (2x)^2}dx$

As we know the integration of this form is $\left [ \because \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a} \right ]$

$= \frac{(\frac{2x}{2})\sqrt{1^2-(2x)^2}+\frac{1^2}{2}\sin^{-1}\frac{2x}{1}}{2\rightarrow Coefficient\ of\ x\ in\ 2x} +C$

$= \frac{1}{2}\left [ x\sqrt{1-4x^2}+\frac{1}{2}\sin^{-1}2x \right ]+C$

$= \frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}2x+C$

Question 3: Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 4x + 6}$

Answer:

Given function $\sqrt{x^2 + 4x + 6}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2 + 4x + 6}dx$

$= \int\sqrt{(x^2 + 4x + 4)+2}dx = \int\sqrt{(x + 2)^2 +(\sqrt2)^2}dx$

And we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

$\Rightarrow I = \frac{x+2}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C$

$= \frac{x+2}{2}\sqrt{x^2+4x+6}+\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C$

Question 4: Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 4x +1}$

Answer:

Given function $\sqrt{x^2 + 4x +1}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2 + 4x + 1}dx$

$= \int\sqrt{(x^2 + 4x + 4)-3}dx = \int\sqrt{(x + 2)^2 -(\sqrt3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$\therefore I = \frac{x+2}{2}\sqrt{x^2+4x+1}-\frac{3}{2}\log\left | (x+2)+\sqrt{x^2+4x+1} \right |+C$

Question 5: Integrate the functions in Exercises 1 to 9.

$\sqrt{1-4x-x^2}$

Answer:

Given function $\sqrt{1-4x-x^2}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1-4x-x^2}dx$

$= \int\sqrt{1-(x^2+4x+4-4)}dx = \int\sqrt{1+4 -(x+2)^2}dx$

$= \int\sqrt{(\sqrt5)^2 -(x+2)^2}dx$

And we know that, $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\therefore I = \frac{x+2}{2}\sqrt{1-4x-x^2}+\frac{5}{2}\sin^{-1}\left ( \frac{x+2}{\sqrt5} \right )+C$

Question 6: Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 4x - 5}$

Answer:

Given function $\sqrt{x^2 + 4x - 5}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2+4x-5}dx$

a $= \int\sqrt{(x^2+4x+4)-9}dx = \int\sqrt{(x+2)^2 -(3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}|+C$

$\therefore I = \frac{x+2}{2}\sqrt{x^2+4x-5}-\frac{9}{2}\log\left | (x+2)+ \sqrt{x^2+4x-5} \right |+C$

Question 7: Integrate the functions in Exercises 1 to 9.

$\sqrt{1 + 3x - x^2}$

Answer:

Given function $\sqrt{1 + 3x - x^2}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1+3x-x^2}dx$

$= \int\sqrt{(1-\left ( x^2-3x+\frac{9}{4}-\frac{9}{4} \right )}dx = \int \sqrt{\left ( 1+\frac{9}{4} \right )-\left ( x-\frac{3}{2} \right )^2}dx$ $= \int \sqrt{\left ( \frac{\sqrt{13}}{2} \right )^2-\left ( x-\frac{3}{2} \right )^2}dx$

And we know that, $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\therefore I = \frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}} \right )+C$

$= \frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{2x-3}{\sqrt{13}} \right )+C$

Question 8: Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 3x}$

Answer:

Given function $\sqrt{x^2 + 3x}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2+3x}dx$

$= \int\sqrt{x^2+3x+\frac{9}{4}-\frac{9}{4}}dx$

$= \int\sqrt{\left ( x+\frac{3}{2} \right )^2-\left ( \frac{3}{2} \right )^2 }dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$\therefore I = \frac{x+\frac{3}{2}}{2}\sqrt{x^2+3x}-\frac{\frac{9}{4}}{2}\log \left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C$

$= \frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\log\left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C$

Question 9: Integrate the functions in Exercises 1 to 9.

$\sqrt{1 + \frac{x^2}{9}}$

Answer:

Given function $\sqrt{1 + \frac{x^2}{9}}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1+\frac{x^2}{9}}dx = \frac{1}{3}\int \sqrt{9+x^2}dx$

$= \frac{1}{3}\int \sqrt{3^2+x^2}dx$

And we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

$\therefore I = \frac{1}{3}\left [ \frac{x}{2}\sqrt{x^2+9} +\frac{9}{2}\log|x+\sqrt{x^2+9}| \right ]+C$

$= \frac{x}{6}\sqrt{x^2+9} +\frac{3}{2}\log\left | x+\sqrt{x^2+9} \right |+C$

Question 10: Choose the correct answer in Exercises 10 to 11.

$\int \sqrt{1+x^2}dx$ is equal to

(A) $\frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log\left |\left(x + \sqrt{1+x^2} \right )\right| +C$

(B) $\frac{2}{3}(1+x^2)^{\frac{3}{2}} + C$

(C) $\frac{2}{3}x(1+x^2)^{\frac{3}{2}} + C$

(D) $\frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log\left |x + \sqrt{1+x^2} \right| +C$

Answer:

As we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

So, $\int \sqrt{1+x^2}dx = \frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log|x+\sqrt{x^2+1}| +C$

Therefore the correct answer is A.

Question 11: Choose the correct answer in Exercises 10 to 11.

$\int \sqrt{x^2 - 8x+7}dx$ is equal to

(A) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} + 9\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C$

(B) $\frac{1}{2}(x+4)\sqrt{x^2-8x+7} + 9\log\left|x+4+\sqrt{x^2 -8x+7}\right| +C$

(C) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} -3\sqrt2\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C$

(D) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} -\frac{9}{2}\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C$

Answer:

Given integral $\int \sqrt{x^2 - 8x+7}dx$

So, let us consider the function to be;

$I = \int\sqrt{x^2-8x+7}dx =\int\sqrt{(x^2-8x+16)-(9)}dx$

$=\int\sqrt{(x-4)^2-(3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$I = \frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2}\log|(x-4)+\sqrt{x^2-8x+7}| +C$

Therefore the correct answer is D.