Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.7

Solution:

This is of the form \( \sqrt{a^2-x^2} \) where \( a=2 \).

Using the formula \( \int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \):

\[ \begin{aligned} \int \sqrt{2^2-x^2} \, dx &= \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} + C \\ &= \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2} + C \end{aligned} \]

Solution:

Rewrite as \( \sqrt{1-(2x)^2} \). Let \( 2x = t \Rightarrow 2dx = dt \Rightarrow dx = \frac{dt}{2} \).

\[ \begin{aligned} \int \sqrt{1-t^2} \frac{dt}{2} &= \frac{1}{2} \left[ \frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin^{-1} t \right] + C \\ &= \frac{1}{2} \left[ \frac{2x}{2}\sqrt{1-4x^2} + \frac{1}{2}\sin^{-1}(2x) \right] + C \\ &= \frac{x}{2}\sqrt{1-4x^2} + \frac{1}{4}\sin^{-1}(2x) + C \end{aligned} \]

Solution:

Complete the square: \( x^2+4x+6 = (x+2)^2 + 2 = (x+2)^2 + (\sqrt{2})^2 \).

Use formula \( \int \sqrt{x^2+a^2} \, dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| + C \).

Here \( X = x+2 \) and \( a = \sqrt{2} \).

\[ \begin{aligned} \int \sqrt{(x+2)^2+(\sqrt{2})^2} \, dx &= \frac{x+2}{2}\sqrt{x^2+4x+6} + \frac{2}{2}\log|x+2+\sqrt{x^2+4x+6}| + C \\ &= \frac{x+2}{2}\sqrt{x^2+4x+6} + \log|x+2+\sqrt{x^2+4x+6}| + C \end{aligned} \]

Solution:

Complete the square: \( x^2+4x+1 = (x+2)^2 - 3 = (x+2)^2 - (\sqrt{3})^2 \).

Use formula \( \int \sqrt{x^2-a^2} \, dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| + C \).

\[ \begin{aligned} \int \sqrt{(x+2)^2-(\sqrt{3})^2} \, dx &= \frac{x+2}{2}\sqrt{x^2+4x+1} - \frac{3}{2}\log|x+2+\sqrt{x^2+4x+1}| + C \end{aligned} \]

Solution:

Complete the square: \( 1-(x^2+4x) = 1-(x^2+4x+4-4) = 1-((x+2)^2-4) = 5-(x+2)^2 \).

Integral is \( \int \sqrt{(\sqrt{5})^2 - (x+2)^2} \, dx \).

Use formula for \( \sqrt{a^2-x^2} \):

\[ \begin{aligned} &= \frac{x+2}{2}\sqrt{1-4x-x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C \end{aligned} \]

Solution:

Complete the square: \( x^2+4x-5 = (x+2)^2 - 9 = (x+2)^2 - 3^2 \).

Use formula for \( \sqrt{x^2-a^2} \):

\[ \begin{aligned} \int \sqrt{(x+2)^2-3^2} \, dx &= \frac{x+2}{2}\sqrt{x^2+4x-5} - \frac{9}{2}\log|x+2+\sqrt{x^2+4x-5}| + C \end{aligned} \]

Solution:

Complete the square: \( 1-(x^2-3x) = 1-(x^2-3x+\frac{9}{4}-\frac{9}{4}) = 1-((x-\frac{3}{2})^2-\frac{13}{4}) = \frac{13}{4}-(x-\frac{3}{2})^2 \).

Integral is \( \int \sqrt{(\frac{\sqrt{13}}{2})^2 - (x-\frac{3}{2})^2} \, dx \).

\[ \begin{aligned} &= \frac{x-3/2}{2}\sqrt{1+3x-x^2} + \frac{13/4}{2}\sin^{-1}\left(\frac{x-3/2}{\sqrt{13}/2}\right) + C \\ &= \frac{2x-3}{4}\sqrt{1+3x-x^2} + \frac{13}{8}\sin^{-1}\left(\frac{2x-3}{\sqrt{13}}\right) + C \end{aligned} \]

Solution:

Complete the square: \( x^2+3x = (x+\frac{3}{2})^2 - \frac{9}{4} \).

Integral is \( \int \sqrt{(x+\frac{3}{2})^2 - (\frac{3}{2})^2} \, dx \).

\[ \begin{aligned} &= \frac{x+3/2}{2}\sqrt{x^2+3x} - \frac{9/4}{2}\log|x+3/2+\sqrt{x^2+3x}| + C \\ &= \frac{2x+3}{4}\sqrt{x^2+3x} - \frac{9}{8}\log|x+\frac{3}{2}+\sqrt{x^2+3x}| + C \end{aligned} \]

Solution:

Rewrite: \( \sqrt{\frac{9+x^2}{9}} = \frac{1}{3}\sqrt{x^2+9} = \frac{1}{3}\sqrt{x^2+3^2} \).

\[ \begin{aligned} \frac{1}{3} \int \sqrt{x^2+3^2} \, dx &= \frac{1}{3} \left[ \frac{x}{2}\sqrt{x^2+9} + \frac{9}{2}\log|x+\sqrt{x^2+9}| \right] + C \\ &= \frac{x}{6}\sqrt{x^2+9} + \frac{3}{2}\log|x+\sqrt{x^2+9}| + C \end{aligned} \]

Answer: (A)

Solution:

Using the standard formula \( \int \sqrt{x^2+a^2} \, dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| + C \) with \( a=1 \).

Result is \( \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log|x+\sqrt{1+x^2}| + C \).

Answer: (D)

Solution:

Complete the square: \( x^2-8x+7 = (x-4)^2 - 16 + 7 = (x-4)^2 - 9 = (x-4)^2 - 3^2 \).

Using formula for \( \sqrt{x^2-a^2} \):

\[ \frac{x-4}{2}\sqrt{x^2-8x+7} - \frac{9}{2}\log|x-4+\sqrt{x^2-8x+7}| + C \]