Exercise 7.7 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download

Ex 7.7 Question 1.

$\sqrt{4-x^2}$

Answer.

$\int \sqrt{4-x^2} d x$
$
\begin{aligned}
& =\int \sqrt{2^2-x^2} d x \\
& =\frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \sin ^{-1} \frac{x}{2}+c \\
& {\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]} \\
& =\frac{x}{2} \sqrt{4-x^2}+2 \sin ^{-1} \frac{x}{2}+c
\end{aligned}
$

Ex 7.7 Question 2.

$\sqrt{1-4 x^2}$
$
\begin{aligned}
& \text { Answer. } \int \sqrt{1-4 x^2} d x \\
& =\int \sqrt{1^2-(2 x)^2} d x \\
& =\frac{\left(\frac{2 x}{2}\right) \sqrt{1^2-(2 x)^2}+\frac{1^2}{2} \sin ^{-1}\left(\frac{2 x}{1}\right)}{2 \rightarrow \text { Coeff. of } x \text { in } 2 x}+c
\end{aligned}
$

$\begin{aligned}
& {\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]} \\
& =\frac{1}{2}\left[x \sqrt{1-4 x^2}+\frac{1}{2} \sin ^{-1} 2 x\right]+c \\
& =\frac{x}{2} \sqrt{1-4 x^2}+\frac{1}{4} \sin ^{-1} 2 x+c \\
& =\frac{1}{4} \sin ^{-1} 2 x+\frac{x}{2} \sqrt{1-4 x^2}+c
\end{aligned}$

Ex 7.7 Question 3.

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

Answer.

$\int \sqrt{x^2+4 x+6} d x=\int \sqrt{x^2+4 x+4+6-4} d x$
$
\begin{aligned}
& =\int \sqrt{(x+2)^2+(\sqrt{2})^2} d x \\
& =\left(\frac{x+2}{2}\right) \sqrt{(x+2)^2+(\sqrt{2})^2}+\frac{(\sqrt{2})^2}{2} \log \left|x+2+\sqrt{(x+2)^2+(\sqrt{2})^2}\right|+c \\
& {\left[\because \int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|\right]} \\
& =\left(\frac{x+2}{2}\right) \sqrt{x^2+4 x+6}+\frac{2}{2} \log \left|x+2+\sqrt{x^2+4 x+6}\right|+c \\
& =\left(\frac{x+2}{2}\right) \sqrt{x^2+4 x+6}+\log \left|x+2+\sqrt{x^2+4 x+6}\right|+c
\end{aligned}
$

Ex 7.7 Question 4.

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

Answer.

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

$\begin{aligned}
& =\int \sqrt{x^2+4 x+4+1-4} d x \\
& =\int \sqrt{(x+2)^2-(\sqrt{3})^2} d x \\
& =\left(\frac{x+2}{2}\right) \sqrt{(x+2)^2-(\sqrt{3})^2}-\frac{(\sqrt{3})^2}{2} \log \left|x+2+\sqrt{(x+2)^2-(\sqrt{3})^2}\right|+c \\
& {\left[\because \int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|\right]} \\
& =\left(\frac{x+2}{2}\right) \sqrt{x^2+4 x+1}-\frac{3}{2} \log \left|x+2+\sqrt{x^2+4 x+1}\right|+c
\end{aligned}$

Ex 7.7 Question 5.

$\sqrt{1-4 x-x^2}$
$
\begin{aligned}
& \text { Ans. } \int \sqrt{1-4 x-x^2} d x \\
& =\int \sqrt{-x^2-4 x+1} d x \\
& =\int \sqrt{-\left(x^2+4 x-1\right)} d x \\
& =\int \sqrt{-\left(x^2+4 x+4-4-1\right)} d x \\
& =\int \sqrt{-\left[(x+2)^2+(\sqrt{5})^2\right]} d x \\
& =\int \sqrt{(\sqrt{5})^2-(x+2)^2} d x \\
& =\frac{x+2}{2} \sqrt{(\sqrt{5})^2-(x+2)^2}+\frac{(\sqrt{5})^2}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+c
\end{aligned}
$

$\begin{aligned}
& {\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]} \\
& =\frac{x+2}{2} \sqrt{1-4 x-x^2}+\frac{5}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+c
\end{aligned}$

Ex 7.7 Question 6.

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

Answer.

$\int \sqrt{x^2+4 x-5} d x$
$
\begin{aligned}
& =\int \sqrt{x^2+4 x+4-4-5} d x \\
& =\int \sqrt{(x+2)^2-(3)^2} d x \\
& =\left(\frac{x+2}{2}\right) \sqrt{(x+2)^2-(3)^2}-\frac{(3)^2}{2} \log \left|x+2+\sqrt{(x+2)^2-(3)^2}\right|+c \\
& {\left[\because \int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|\right]} \\
& =\left(\frac{x+2}{2}\right) \sqrt{x^2+4 x-5}-\frac{9}{2} \log \left|x+2+\sqrt{x^2+4 x-5}\right|+c
\end{aligned}
$

Ex 7.7 Question 7.

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

Answer.

$\int \sqrt{1+3 x-x^2} d x$
$
\begin{aligned}
& =\int \sqrt{-x^2+3 x+1} d x \\
& =\int \sqrt{-\left(x^2-3 x-1\right)} d x \\
& =\int \sqrt{-\left(x^2-3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-1\right)} d x
\end{aligned}
$

$\begin{aligned}
& =\int \sqrt{\left[\left(x-\frac{3}{2}\right)^2+\left(\frac{\sqrt{13}}{2}\right)^2\right]} d x \\
& =\int \sqrt{\left(\frac{\sqrt{13}}{1}\right)^2-\left(x-\frac{3}{2}\right)^2} d x \\
& =\frac{x-\frac{3}{2}}{2} \sqrt{\left(\frac{\sqrt{13}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}+\frac{\left(\frac{\sqrt{13}}{2}\right)^2}{2} \sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}\right)+c \\
& {\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]} \\
& =\left(\frac{2 x-3}{4}\right) \sqrt{1+3 x-x^2}+\frac{13}{8} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{13}}\right)+c
\end{aligned}$

Ex 7.7 Question 8.

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

Answer.

$\int \sqrt{x^2+3 x} d x$
$
\begin{aligned}
& =\int \sqrt{x^2+3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2} d x \\
& =\int \sqrt{\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2} d x \\
& =\frac{x+\frac{3}{2}}{2} \sqrt{\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2}-\frac{\left(\frac{3}{2}\right)^2}{2} \log \left|x+\frac{3}{2}+\sqrt{\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2}\right|+c
\end{aligned}
$

$\begin{aligned}
& {\left[\because \int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|\right]} \\
& =\frac{2 x+3}{4} \sqrt{x^2+3 x}-\frac{9}{8} \log \left|x+\frac{3}{2}+\sqrt{x^2+3 x}\right|+c
\end{aligned}$

Ex 7.7 Question 9.

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

Answer.

$\int \sqrt{1+\frac{x^2}{9}} d x$
$
\begin{aligned}
& =\int \sqrt{\frac{9+x^2}{9}} d x \\
& =\int \frac{\sqrt{x^2+3^2}}{3} d x \\
& =\frac{1}{3} \int \sqrt{x^2+3^2} d x \\
& =\frac{1}{3}\left[\frac{x}{2} \sqrt{x^2+3^2}+\frac{3^2}{2} \log \left|x+\sqrt{x^2+3^2}\right|\right]+c \\
& {\left[\because \int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|\right]+c} \\
& =\frac{x}{6} \sqrt{x^2+9}+\frac{3}{2} \log \left|x+\sqrt{x^2+9}\right|+c
\end{aligned}
$

Ex 7.7 Question 10.

$\text {} \int \sqrt{1+x^2} d x \text { is equal to: }$

(A) $\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^2}\right)\right|+\mathrm{C}$
(B) $\frac{2}{3}\left(1+x^2\right)^{\frac{3}{2}}+\mathrm{C}$
(C) $\frac{2}{3} x\left(1+x^2\right)^{\frac{3}{2}}+\mathrm{C}$
(D) $\frac{x^2}{2} \sqrt{1+x^2}+\frac{1}{2} x^2 \log \left|\left(x+\sqrt{1+x^2}\right)\right|+\mathrm{C}$
$
\begin{aligned}
& \text { Answer. } \int \sqrt{1+x^2} d x \\
& =\int \sqrt{x^2+1^2} d x \\
& =\frac{x}{2} \sqrt{x^2+1^2}+\frac{1^2}{2} \log \left|x+\sqrt{x^2+1^2}\right|+\mathrm{C} \\
& =\frac{x}{2} \sqrt{x^2+1}+\frac{1}{2} \log \left|x+\sqrt{x^2+1}\right|+\mathrm{C} \\
& {\left[\because \int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|\right]+c}
\end{aligned}
$

Therefore, option (A) is correct.

Ex 7.7 Question 11.

$\int \sqrt{x^2-8 x+7} d x$ is equal to
(A) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}+9 \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$
(B) $\frac{1}{2}(x+4) \sqrt{x^2-8 x+7}+9 \log \left|x+4+\sqrt{x^2-8 x+7}\right|+C$
(C) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$
(D) $\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C$

Answer.

$\int \sqrt{x^2-8 x+7} d x$
$
\begin{aligned}
& =\int \sqrt{x^2-8 x+16-16+7} d x \\
& =\int \sqrt{(x-4)^2-3^2} d x \\
& =\left(\frac{x-4}{2}\right) \sqrt{(x-4)^2-3^2}-\frac{3^2}{2} \log \left|x-4+\sqrt{(x-4)^2-3^2}\right|+\mathrm{C} \\
& {\left[\because \int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|\right]} \\
& =\left(\frac{x-4}{2}\right) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+\mathrm{C}
\end{aligned}
$

Therefore, option (D) is correct.