Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.7

In the article, we will solve Exercise 7.7 from Chapter 7, “Integrals” in the NCERT. Exercise 7.7 covers some special types of standard integrals which are based on integration by parts

Integral Formulae To Solve Exercise 7.7

\int \sqrt{a^2-x^2}dx \space = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})+c...(i)

\int \sqrt{x^2+a^2}dx \space = \frac{x}{2} \sqrt{x^2+a^2} + \frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|+c...(iI)

\int \sqrt{x^2-a^2}dx \space = \frac{x}{2} \sqrt{x^2-a^2} - \frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|+c...(iII)

Exercise 7.7 Solutions

Q.1: \int \sqrt{1-4x^2}dx

Solution:

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

by putting a = 2 in formula (1), we get

=\frac{x}{2} \sqrt{2^2-x^2} + \frac{2^2}{2}\sin^{-1}(\frac{x}{2})+c

=\frac{x}{2} \sqrt{4-x^2} + 2\sin^{-1}(\frac{x}{2})+c

Q.2: \int \sqrt{1-4x^2}dx

Solution:

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

by putting 2x = y, 2dx = dy, dx = \frac{dy}{2}

=\frac{1}{2}\int \sqrt{1-y^2}dy

now by using formula (1), we get

=\frac{1}{2}[\frac{y}{2} \sqrt{1-y^2} + \frac{1^2}{2}\sin^{-1}(\frac{y}{1})]+c

now we put y = 2x

=\frac{1}{2}[\frac{2x}{2} \sqrt{1-(2x)^2} + \frac{1^2}{2}\sin^{-1}(2x)]+c

=\frac{x}{2} \sqrt{1-4x^2} + \frac{1}{4}\sin^{-1}(2x)+c

Q.3: \int \sqrt{x^2+4x+6}dx

Solution:

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

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

let (x+2) = y and dx = dy

\int \sqrt{y^2+(\sqrt2)^2}dy

using formula (2) , we get

= \frac{y}{2} \sqrt{y^2+(\sqrt2)^2} + \frac{ (\sqrt2)^2}{2}\ln|y+\sqrt{y^2+(\sqrt2)^2}|+c

now putting y = (x+2)

= \frac{(x+2)}{2} \sqrt{(x+2)^2+(\sqrt2)^2} + \ln|x+2+\sqrt{(x+2)^2+(\sqrt2)^2}|+c

= \frac{(x+2)}{2} \sqrt{x^2+4x+6} + \ln|x+2+\sqrt{x^2+4x+6}|+c

Q.4: \int \sqrt{x^2+4x+1}dx

Solution:

\int \sqrt{x^2+4x+1}dx \space = \int \sqrt{x^2+4x+4-4+1}dx

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

substitute (x+2) = y and dx = dy

= \int \sqrt{y^2-(\sqrt3)^2}dy \space = \frac{y}{2} \sqrt{y^2-(\sqrt3)^2} - \frac{(\sqrt3)^2}{2}\ln|y+\sqrt{y^2-(\sqrt3)^2}|+c

= \frac{(x+2)}{2} \sqrt{(x+2)^2-(\sqrt3)^2} - \frac{3}{2}\ln|x+2+\sqrt{(x+2)^2-(\sqrt3)^2}|+c

= \frac{(x+2)}{2} \sqrt{x^2+4x+1} - \frac{3}{2}\ln|x+2+\sqrt{x^2+4x+1}|+c

Q.5: \int \sqrt{1-4x-x^2}dx

Solution:

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

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

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

now putting (x+2) = y and dx = dy

\int \sqrt{(\sqrt5)^2-y^2}dy

using formula (1)

= \frac{y}{2} \sqrt{(\sqrt5)^2-y^2} + \frac{(\sqrt5)^2}{2}\sin^{-1}(\frac{y}{\sqrt5})+c

now putting (x+2) =y

= \frac{(x+2)}{2} \sqrt{(\sqrt5)^2-(x+2)^2} + \frac{5}{2}\sin^{-1}(\frac{x+2}{\sqrt5})+c

= \frac{(x+2)}{2} \sqrt{1-4x-x^2} + \frac{5}{2}\sin^{-1}(\frac{x+2}{\sqrt5})+c

Q.6: \int \sqrt{x^2+4x-5}dx

Solution:

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

\int \sqrt{(x+2)^2-3^2}dx

now substitute (x+2) = y and dx = dy

\int \sqrt{y^2-3^2}dy \space = \frac{y}{2} \sqrt{y^2-3^2} - \frac{3^2}{2}\ln|y+\sqrt{y^2-3^2}|+c

now put (x+2) = y

= \frac{(x+2)}{2} \sqrt{(x+2)^2-3^2} - \frac{9}{2}\ln|x+2+\sqrt{(x+2)^2-3^2}|+c

= \frac{(x+2)}{2} \sqrt{x^2+4x-5} - \frac{9}{2}\ln|x+2+\sqrt{x^2+4x-5}|+c

Q.7: \int \sqrt{1+3x-x^2}dx

Solution:

\int \sqrt{1+3x-x^2}dx \space = \int \sqrt{-(x^2-3x-1)}dx

= \int \sqrt{-(x^2-3x+(\frac{3}{2})^2-(\frac{3}{2})^2-1)}dx

= \int \sqrt{-((x-\frac{3}{2})^2-(\frac{\sqrt13}{2})^2)}dx

= \int \sqrt{(\frac{\sqrt13}{2})^2-(x-\frac{3}{2})^2}dx

now substitute , (x-\frac{3}{2}) = y and dx = dy

= \int \sqrt{(\frac{\sqrt13}{2})^2-y^2}dy

using formula (1), we get

= \frac{y}{2} \sqrt{(\frac{\sqrt13}{2})^2-y^2} + \frac{(\frac{\sqrt13}{2})^2}{2}\sin^{-1}(\frac{y}{\frac{\sqrt13}{2}})+c

= \frac{(x-\frac{3}{2})}{2} \sqrt{(\frac{\sqrt13}{2})^2-(x-\frac{3}{2})^2} + \frac{13}{8}\sin^{-1}(\frac{(x-\frac{3}{2})}{\frac{\sqrt13}{2}})+c

= \frac{(2x-3)}{4} \sqrt{1+3x-x^2} + \frac{13}{8}\sin^{-1}(\frac{2x-3}{\sqrt13})+c

Q.8: \int \sqrt{x^2+3x}dx

Solution:

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

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

now substitute (x+\frac{3}{2}) = y and dx = dy

= \int \sqrt{y^2 - (\frac{3}{2})^2}dy

using formula (3) , we get

= \frac{y}{2} \sqrt{y^2-(\frac{3}{2})^2} - \frac{(\frac{3}{2})^2}{2}\ln|y+\sqrt{y^2-(\frac{3}{2})^2}|+c

= \frac{(x+\frac{3}{2})}{2} \sqrt{(x+\frac{3}{2})^2-(\frac{3}{2})^2} - \frac{9}{2}\ln|x+\frac{3}{2}+\sqrt{(x+\frac{3}{2})^2-(\frac{3}{2})^2}|+c

= \frac{(2x+3)}{4} \sqrt{x^2+3x} - \frac{9}{8}\ln|x+\frac{3}{2}+\sqrt{x^2+3x}|+c

Q.9: \int \sqrt{1+(\frac{x^2}{9})}dx

Solution:

\int \sqrt{1+(\frac{x^2}{9})}dx = \frac{1}{3}\int \sqrt{x^2+3^2)}dx

by using formula (2)

\frac{1}{3}\int \sqrt{x^2+3^2}dx \space = \frac{1}{3}[\frac{x}{2} \sqrt{x^2+3^2} + \frac{3^2}{2}\ln|x+\sqrt{x^2+3^2}|]+c

= \frac{x}{6} \sqrt{x^2+9} + \frac{3}{2}\ln|x+\sqrt{x^2+9}|+c

Q.10: \int \sqrt{1+x^2}dx

(A) x/2√(1 + x²) + 1/2log|x + √(1 + x²) + C|

(B) 2/3(1 + x²)^3/2 + C

(C) 2/3x√(1 + x²)^3/2 + C

(D) x²/2√(1 + x²) + 1/2.x².log|x + √(1 + x²) + C|

Solution:

Option (A) is Correct

Using formula (2) we get

\int \sqrt{x^2+1^2}dx \space = \frac{x}{2} \sqrt{x^2+1^2} + \frac{1^2}{2}\ln|x+\sqrt{x^2+1^2}|+c

= \frac{x}{2} \sqrt{x^2+1} + \frac{1}{2}\ln|x+\sqrt{x^2+1}|+c

Q.11: \int \sqrt{x^2-8x+7}dx

(A) 1/2(x - 4)√(x² - 8x + 7) + 9.log|x - 4 + √(x² -8x + 7)| + C

(B) 1/2(x - 4)√(x² - 8x + 7) + 9.log|x + 4 + √(x² -8x + 7)| + C

(C) 1/2(x - 4)√(x² - 8x + 7) - 3√(2).log|x - 4 + √(x² -8x + 7)| + C

(D) 1/2(x - 4)√(x² - 8x + 7) - 9/2.log|x - 4 + √(x² -8x + 7)| + C

Solution:

Option (D) is Correct

\int \sqrt{x^2-8x+7}dx \space = \int \sqrt{x^2-8x+4^2-4^2+7}dx

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

let (x-4) = y and dx = dy

\int\sqrt{y^2-3^2}dy

using formula (3)

\int \sqrt{y^2-3^2}dy \space = \frac{y}{2} \sqrt{y^2-3^2} - \frac{3^2}{2}\ln|y+\sqrt{y^2-3^2}|+c

Now put y = (x-4) and after simplifying

= \frac{(x-4)}{2} \sqrt{x^2-8x+7} - \frac{9}{2}\ln|x-4+\sqrt{x^2-8x+7}|+c