Ex 7.1 Class 12 Maths Ncert Solutions

Method: To find the anti-derivative of a function by inspection
Steps. 1. In this method we look for a function whose derivative is the given function. For Example. if we need to find anti derivative of \( x^{2} \), we know that derivative of \( x^{3} \) is \( 3 x^{2} \). Therefore, the variable terms come out to be the same.
2. After that balance out the coefficients of variables by dividing and multiplying suitable terms. From above example if [we divide \( x^{3} \) by 3 we will get the answer as \( x^{2} \). Hence, we can say that anti derivative of \( x^{2} \) is \( \frac{ x^{3} }{ 3 } \).
Now, similarly, we know that \( \frac{\mathrm{d}(\cos 2 x)}{\mathrm{dx}}=-2 \sin 2 x \)
\(\Rightarrow \sin 2 x=-\frac{1}{2}\cdot \frac{\mathrm{d}(\cos 2 x)}{\mathrm{dx}}\)
\(
=\frac{d}{d x}\left(-\frac{1}{2} \cos 2 x\right)
\)
Therefore, the anti-derivative of \( \sin 2 x \) is \( -\frac{1}{2} \cos 2 x \)

We know that \( \frac{d(\sin 3 x)}{d x}=3 \cos 3 x \)
\(
\Rightarrow \cos 3 x=\frac{1}{3} \cdot\frac{d(\sin 3 x)}{d x}\)
\(
=\frac{d}{d x}\left(\frac{1}{3} \sin 3 x\right)
\)
Therefore, the anti-derivative of \( \cos 3 x \) is \( \frac{1}{3} \sin 3 x \)

We know that \( \frac{d\left(e^{2 x}\right)}{d x}=2 e^{2 x} \)
\(
\Rightarrow \mathrm{e}^{2 x}=\frac{1}{2} \cdot \frac{\mathrm{d}\left(\mathrm{e}^{2 x}\right)}{\mathrm{d}x}\)
\(
=\frac{d}{d x}\left(\frac{1}{2} \mathrm{e}^{2 x}\right)
\)
Therefore, the anti-derivative of \( \mathrm{e}^{2 x} \) is \( \frac{1}{2} \mathrm{e}^{2 x} \).

We know that \( \frac{d}{d x}(a x+b)^{3}=3 a(a x+b)^{2} \)
\(
\Rightarrow(ax+\mathrm{b})^{2}=\frac{1}{3 \mathrm{a}} \cdot \frac{\mathrm{d}(ax+\mathrm{b})^{3}}{\mathrm{dx}}\)
\(
=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{3 \mathrm{a}}(ax+\mathrm{b})^{3}\right)
\)
Therefore, the anti-derivative of \( (ax+\mathrm{b})^{2} \) is \( \frac{1}{3 a}(ax+\mathrm{b})^{3} \).

We know that
\(
\frac{d}{d x} \cos 2 x=-2 \sin 2 x\)
\(
\Rightarrow-\frac{1}{2} \cdot \frac{d}{d x}(\cos 2 x)=\sin 2 x\)
\(
\Rightarrow \frac{d}{d x}\left(-\frac{1}{2} \cos 2 x\right)=\sin 2 x
\)
Therefore, the anti-derivative of \( \sin 2x \) is \( -\frac{1}{2} \cos 2 x\quad \ldots (1)\)
Also,
\(\frac{d}{d x} e^{3 x}=3 e^{3 x}\)
\(\Rightarrow \frac{4}{3}\cdot \frac{d}{d x} e^{3 x}=3 e^{3 x}\)
Therefore, the anti-derivative of \( 4 \mathrm{e}^{3 x} \) is \( \frac{4}{3} e^{3 x} \quad\ldots (2)\).
From (1) and (2), we get,
\(\frac{d}{d x}\left(-\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\right)\)
\(=\sin 2 x-4 \mathrm{e}^{3 x}\)
Therefore, the anti-derivative of \( \sin 2 x-4 \ e^{3 x} \) is
\(-\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\)

\(\int\left(4 ~e^{3 x}+1\right) d x=\int e^{3 x} d x+\int 1 d x\)
\(=4\left(\frac{e^{3 x}}{3}\right)+x+C\)
\(=\frac{4}{3} e^{3 x}+x+C\)

\(
\int x^{2}\left(1-\frac{1}{x^{2}}\right) d x\)
\(
=\int\left(x^{2}-1\right) d x\)
\(
=\int x^{2} d x-\int 1 d x
\)
\(=\frac{x^{3}}{x}-x+C\)

\(\int\left(a x^{2}+b x+c\right) d x\)
\(=a \int x^{2} d x+b \int x d x+c \int 1 d x\)
\(=a\left(\frac{x^{3}}{3}\right)+b\left(\frac{x^{2}}{2}\right)+c x+C\)
\(=\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+C\)

\(\int 2 x^{2}+e^{x} d x\)
\(=2 \int x^{2} d x+\int e^{x} 1 d x\)
\(=2\left(\frac{x^{3}}{3}\right)+e^{x}+C\)
\(=\frac{2 x^{3}}{3}+e^{x}+C\)

\(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x\)
\(=\int\left(x+\frac{1}{x}-2\right) d x\)
\(=\int x d x+\int \frac{1}{x} d x-2 \int 1 . d x\)
Now we know that,
\(\int x^{n} d x=\frac{x^{n}+1}{n+1}+c\)
Therefore,
\(=\frac{x^{2}}{2}+\log |x|-2 x+C\)

\(\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x\)
Separating the terms, we get,
\(=\int\left(x+5-4 x^{-2}\right) d x\)
Applying the formula,
\(\int x^{n} d x=\frac{x^{n}+1}{n+1}+c\)
\(=\int x d x+5 \int 1 . d x-4 \int x^{-2} \cdot d x\)

\(\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x\)
Separating the terms, we get,
\(=\int\left(x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}\right) d x\)
Applying the formula,
\(\int x^{n} d x=\frac{x^{n}+1}{n+1}+c\)
\(=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left(x^{\frac{3}{2}}\right)}{\frac{3}{2}}+\frac{4\left(x^{\frac{1}{2}}\right)}{\frac{1}{2}}+C\)
\(=\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 x^{\frac{1}{2}}+C\)
\(=\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+C\)

\(\int \frac{x^{3}-x^{2}+x-1}{x-1} d x\)
Now the numerator can be factorized as,
\(x^{3}-x^{2}+x-1=x^{2}(x-1)+1(x-1)\)
\(x^{3}-x^{2}+x-1=\left(x^{2}+1\right)(x-1)\)
Now putting this in given integral we get,
\(\frac{x^{3}-x^{2}+x-1}{x-1}=\frac{\left(x^{2}+1\right)(x-1)}{x-1}=x^{2}+1\)
\(=\int\left(x^{2}+1\right) d x\)
\(=\int x^{2} d x+\int 1 \cdot d x\)
\(=\frac{x^{3}}{3}+x+C\)

\(\int(1-x) \sqrt{x} ~d x\)
\(=\int\left(\sqrt{x}-x^{\frac{3}{2}}\right) d x\)
\(=\int x^{\frac{1}{2}} d x-\int x^{\frac{3}{2}} d x\)
\(=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+C\)
\(=\frac{2}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}+C\)

\(\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x\)
\(=\int\left(3 x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+3 x^{\frac{1}{2}}\right) d x\)
\(=3\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+2\left(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right)+3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\)
\(=\frac{6}{7} x^{\frac{7}{2}}+\frac{4}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+\mathrm{C}\)

\(\int\left(2 x-3 \cos x+e^{x}\right) d x\)
\(=2 \int x d x-3 \int \cos x d x+\int e^{x} d x\)
\(=\frac{2 x^{2}}{2}-3(\sin x)+e^{x}+C\)
\(=x^{2}-3 \sin x+e^{x}+C\)

\(\int\left(2 x^{2}-3 \sin x+5 \sqrt{x}\right) d x\)
\(=2 \int x^{2} d x-3 \int \sin d x+5 \int x^{\frac{1}{2}} d x\)
\(=\frac{2 x^{3}}{3}+3 \cos x+5\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\)
\(=\frac{2 x^{3}}{3}+3 \cos x+\frac{10}{3} x^{\frac{3}{2}}+C\)

Formulas Used:
\( \int \sec ^{2} x \mathrm{~d} x=\tan x+\mathrm{c} \) and \( \int \sec x \tan x \mathrm{d}x=\sec x+\mathrm{c} \)
\(\int \sec x(\sec x+\tan x) d x\)
Opening the brackets, we get,
\(=\int\left(\sec ^{2} x \mathrm{~d} x+\sec x \tan x\right) \mathrm{d}x\)
\(=\int \sec ^{2} x \mathrm{~d} x+\int \sec x \tan x \mathrm{d}x\)
\(=\tan x+\sec x+\mathrm{C}\)

\(\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x\)
\(=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\sin ^{2} x}} d x\)
\(=\int \frac{\sin ^{2} x}{\cos ^{2} x} d x\)
\(=\int \tan ^{2} x d x\)
\(=\int\left(\sec ^{2} x-1\right) d x\)
\(=\int \sec ^{2} x d x-\int 1 . d x\)
\(=\tan x-x+C\)

\(\int \frac{2-3 \sin x}{\cos ^{2} x} d x\)
\(=\int\left(\frac{2}{\cos ^{2} x}-\frac{3 \sin x}{\cos ^{2} x}\right) d x\)
\(=\int 2 \sec ^{2} x d x-3 \int \tan x \sec x d x\)
\(=2 \tan x-3 \sec x+\mathrm{C}\)

\(\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x\)
\(=\int x^{\frac{1}{2}} d x+\int x^{-\frac{1}{2}} d x\)
\(=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C\)
\(=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)

It is given that \( \frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}} \)
\(f(x)=\int 4 x^{3}-\frac{3}{x^{4}} d x\)
\(
\Rightarrow f(x)=4 \int x^{3} d x-3 \int\left(x^{-4}\right) d x\)
\(
\Rightarrow f(x)=4\left(\frac{x^{4}}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C\)
\(
\Rightarrow f(x)=x^{4}+\frac{1}{x^{3}}+C
\)
Also, it is given that \( f(2)=0 \)
\(\Rightarrow f(2)=(2)^{4}+\frac{1}{(2)^{3}}+C=0\)
\(\Rightarrow 16+\frac{1}{8}+C=0\)
\(\Rightarrow C=-\left(16+\frac{1}{8}\right)\)
\(\Rightarrow C=-\frac{129}{8}\)
Therefore, \( f(x)=x^{4}+\frac{1}{x^{3}}-\frac{129}{8} \)