Exercise 7.1 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths
Updated On 26-08-2025 By Lithanya
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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download
Ex 7.1 Question 1:
$\sin 2 x$
Answer:
The anti derivative of $\sin 2 x$ is a function of $x$ whose derivative is $\sin 2 x$.
It is known that,
$
\begin{aligned}
& \frac{d}{d x}(\cos 2 x)=-2 \sin 2 x \\
& \Rightarrow \sin 2 x=-\frac{1}{2} \frac{d}{d x}(\cos 2 x) \\
& \therefore \sin 2 x=\frac{d}{d x}\left(-\frac{1}{2} \cos 2 x\right)
\end{aligned}
$
Therefore, the anti derivative of
$
\sin 2 x \text { is }-\frac{1}{2} \cos 2 x
$
Ex 7.1 Question 2:
$\operatorname{Cos} 3 x$
Answer:
The anti derivative of $\cos 3 x$ is a function of $x$ whose derivative is $\cos 3 x$.
It is known that,
$
\begin{aligned}
& \frac{d}{d x}(\sin 3 x)=3 \cos 3 x \\
& \Rightarrow \cos 3 x=\frac{1}{3} \frac{d}{d x}(\sin 3 x) \\
& \therefore \cos 3 x=\frac{d}{d x}\left(\frac{1}{3} \sin 3 x\right)
\end{aligned}
$
Therefore, the anti derivative of
$\cos 3 x \text { is } \frac{1}{3} \sin 3 x$
Ex 7.1 Question 3:
$e^{2 x}$
Answer:
The anti derivative of $e^{2 x}$ is the function of $x$ whose derivative is $\mathrm{e}^{2 x}$.
It is known that,
$
\begin{aligned}
& \frac{d}{d x}\left(e^{2 x}\right)=2 e^{2 x} \\
& \Rightarrow e^{2 x}=\frac{1}{2} \frac{d}{d x}\left(e^{2 x}\right) \\
& \therefore e^{2 x}=\frac{d}{d x}\left(\frac{1}{2} e^{2 x}\right)
\end{aligned}
$
Therefore, the anti derivative of
$
e^{2 x} \text { is } \frac{1}{2} e^{2 x}
$
Ex 7.1 Question 4:
$
(a x+b)^2
$
Answer:
The anti derivative of $(a x+b)^2$ is the function of $x$ whose derivative is $(a x+b)^2$. It is known that,
$
\begin{aligned}
& \frac{d}{d x}(a x+b)^3=3 a(a x+b)^2 \\
& \Rightarrow(a x+b)^2=\frac{1}{3 a} \frac{d}{d x}(a x+b)^3 \\
& \therefore(a x+b)^2=\frac{d}{d x}\left(\frac{1}{3 a}(a x+b)^3\right)
\end{aligned}
$
Therefore, the anti derivative of $(a x+b)^2$ is $\frac{1}{3 a}(a x+b)^3$
Ex 7.1 Question 5:
$
\sin 2 x-4 e^{3 x}
$
Answer:
The anti derivative of $\left(\sin 2 x-4 e^{3 x}\right)$ is the function of $x$ whose derivative is $\left(\sin 2 x-4 e^{3 x}\right)$.
It is known that,
$
\frac{d}{d x}\left(-\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\right)=\sin 2 x-4 e^{3 x}
$
Therefore, the anti derivative of $\left(\sin 2 x-4 e^{3 x}\right)$ is $\left(-\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\right)$.
Ex 7.1 Question 6:
$
\int\left(4 e^{3 x}+1\right) d x
$
Answer:
$
\begin{aligned}
& \int\left(4 e^{3 x}+1\right) d x \\
& =4 \int e^{3 x} d x+\int 1 d x \\
& =4\left(\frac{e^{3 x}}{3}\right)+x+\mathrm{C} \\
& =\frac{4}{3} e^{3 x}+x+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 7:
$
\int x^2\left(1-\frac{1}{x^2}\right) d x
$
Answer:
$
\begin{aligned}
& \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}{3}-x+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 8:
$
\int\left(a x^2+b x+c\right) d x
$
Answer:
$
\begin{aligned}
& \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 \cdot d x \\
& =a\left(\frac{x^3}{3}\right)+b\left(\frac{x^2}{2}\right)+c x+\mathrm{C} \\
& =\frac{a x^3}{3}+\frac{b x^2}{2}+c x+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 9:
$
\int\left(2 x^2+e^x\right) d x
$
Answer:
$
\int\left(2 x^2+e^x\right) d x
$
$
\begin{aligned}
& =2 \int x^2 d x+\int e^x d x \\
& =2\left(\frac{x^3}{3}\right)+e^x+\mathrm{C} \\
& =\frac{2}{3} x^3+e^x+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 10:
$
\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 d x
$
Answer:
$
\begin{aligned}
& \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 \cdot d x \\
& =\frac{x^2}{2}+\log |x|-2 x+\text { C }
\end{aligned}
$
Ex 7.1 Question 11:
$
\int \frac{x^3+5 x^2-4}{x^2} d x
$
Answer:
$
\int \frac{x^3+5 x^2-4}{x^2} d x
$
$
\begin{aligned}
& =\int\left(x+5-4 x^{-2}\right) d x \\
& =\int x d x+5 \int 1 \cdot d x-4 \int x^{-2} d x \\
& =\frac{x^2}{2}+5 x-4\left(\frac{x^{-1}}{-1}\right)+\mathrm{C} \\
& =\frac{x^2}{2}+5 x+\frac{4}{x}+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 12:
$
\int \frac{x^3+3 x+4}{\sqrt{x}} d x
$
Answer:
$
\begin{aligned}
& \int \frac{x^3+3 x+4}{\sqrt{x}} d x \\
& =\int\left(x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}\right) d x \\
& =\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}}+\mathrm{C} \\
& =\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 x^{\frac{1}{2}}+\mathrm{C} \\
& =\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+\mathrm{C} \\
&
\end{aligned}
$
Ex 7.1 Question 13:
$
\int \frac{x^3-x^2+x-1}{x-1} d x
$
Answer:
$
\int \frac{x^3-x^2+x-1}{x-1} d x
$
On dividing, we obtain
$
\begin{aligned}
& =\int\left(x^2+1\right) d x \\
& =\int x^2 d x+\int 1 d x \\
& =\frac{x^3}{3}+x+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 14:
$
\int(1-x) \sqrt{x} d x
$
Answer:
$
\begin{aligned}
& \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}}+\mathrm{C} \\
= & \frac{2}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 15:
$
\int \sqrt{x}\left(3 x^2+2 x+3\right) d x
$
Answer:
$
\int \sqrt{x}\left(3 x^2+2 x+3\right) d x
$
$\begin{aligned}
& =\int\left(3 x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+3 x^{\frac{1}{2}}\right) d x \\
& =3 \int x^{\frac{5}{2}} d x+2 \int x^{\frac{3}{2}} d x+3 \int x^{\frac{1}{2}} d x \\
& =3\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+2\left(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right)+3 \frac{\left(x^{\frac{3}{2}}\right)}{\frac{3}{2}}+\mathrm{C} \\
& =\frac{6}{7} x^{\frac{7}{2}}+\frac{4}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+\mathrm{C}
\end{aligned}$
Ex 7.1 Question 16:
$
\int\left(2 x-3 \cos x+e^x\right) d x
$
Answer:
$
\begin{aligned}
& \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+\mathrm{C} \\
& =x^2-3 \sin x+e^x+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 17:
$
\int\left(2 x^2-3 \sin x+5 \sqrt{x}\right) d x
$
Answer:
$
\int\left(2 x^2-3 \sin x+5 \sqrt{x}\right) d x
$
$
\begin{aligned}
& =2 \int x^2 d x-3 \int \sin x 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)+\mathrm{C} \\
& =\frac{2}{3} x^3+3 \cos x+\frac{10}{3} x^{\frac{3}{2}}+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 18:
$
\int \sec x(\sec x+\tan x) d x
$
Answer:
$
\begin{aligned}
& \int \sec x(\sec x+\tan x) d x \\
& =\int\left(\sec ^2 x+\sec x \tan x\right) d x \\
& =\int \sec ^2 x d x+\int \sec x \tan x d x \\
& =\tan x+\sec x+\mathrm{C}
\end{aligned}
$
Ex 7.1 Question 19:
$
\int \frac{\sec ^2 x}{\operatorname{cosec}^2 x} d x
$
Answer:
$
\int \frac{\sec ^2 x}{\operatorname{cosec}^2 x} d x
$
$\begin{aligned}
& =\int \frac{\frac{1}{\cos ^2 x}}{\frac{1}{\sin ^2 x}} d x \\
& =\int \frac{\sin ^2 x}{\cos ^2 x} d x \\
& =\int \tan ^2 x d x \\
& =\int\left(\sec ^2 x-1\right) d x \\
& =\int \sec ^2 x d x-\int 1 d x \\
& =\tan x-x+C
\end{aligned}$
Ex 7.1 Question 20:
$
\int \frac{2-3 \sin x}{\cos ^2 x} d x
$
Answer:
$
\begin{aligned}
& \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}
\end{aligned}
$
Ex 7.1 Question 21:
The anti derivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals
(A) $\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+C$
(B) $\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^2+\mathrm{C}$
(C) $\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+\mathrm{C}$
(D) $\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+\mathrm{C}$
Answer:
$
\begin{aligned}
& \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}}+\mathrm{C} \\
& =\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+\mathrm{C}
\end{aligned}
$
Hence, the correct answer is C.
Ex 7.1 Question 22:
If $\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4}$ such that $f(2)=0$, then $f(x)$ is
(A) $x^4+\frac{1}{x^3}-\frac{129}{8}$
(B) $x^3+\frac{1}{x^4}+\frac{129}{8}$
(C) $x^4+\frac{1}{x^3}+\frac{129}{8}$
(D) $x^3+\frac{1}{x^4}-\frac{129}{8}$
Answer:
It is given that,
$
\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4}
$
$\therefore$ Anti derivative of
$
4 x^3-\frac{3}{x^4}=f(x)
$
$
\begin{aligned}
& \therefore f(x)=\int 4 x^3-\frac{3}{x^4} d x \\
& f(x)=4 \int x^3 d x-3 \int\left(x^{-4}\right) d x \\
& f(x)=4\left(\frac{x^4}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+\mathrm{C} \\
& f(x)=x^4+\frac{1}{x^3}+\mathrm{C}
\end{aligned}
$
Also,
$
\begin{aligned}
& f(2)=0 \\
& \therefore f(2)=(2)^4+\frac{1}{(2)^3}+\mathrm{C}=0 \\
& \Rightarrow 16+\frac{1}{8}+\mathrm{C}=0 \\
& \Rightarrow \mathrm{C}=-\left(16+\frac{1}{8}\right) \\
& \Rightarrow \mathrm{C}=\frac{-129}{8} \\
& \therefore f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}
\end{aligned}
$
Hence, the correct answer is $\mathrm{A}$.
