Anand Classes offers a complete and free-downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 7 – Integrals, Exercise 7.1 (Set-1), created by expert educators to guide you through the fundamental techniques of integration, such as finding antiderivatives, applying linear operators of integrals, and breaking down complex integrals using properties of integrals as inverse of differentiation. These solutions align perfectly with the latest CBSE / NCERT syllabus and are crafted to support solid preparation for board exams and competitive tests. Click the print button to download study material and notes.
NCERT Question.1 : Finding the Anti-Derivative (Integral) by Inspection of function $\sin 2x$.
Solution :
Anti-derivative of $\sin 2x$ is a function of $x$ whose derivative is $\sin 2x$.
It is known that,
$$\frac{d}{dx}(\cos 2x) = -2 \sin 2x$$
Therefore,
$$\sin 2x = -\frac{1}{2} \frac{d}{dx}(\cos 2x)$$
or
$$\sin 2x = \frac{d}{dx}\left(-\frac{1}{2} \cos 2x\right)$$
Hence, the anti-derivative of $\sin 2x$ is
$$-\frac{1}{2} \cos 2x + C$$
NCERT Question.2 : Finding the Anti-Derivative (Integral) by Inspection of function $\cos 3x$.
Solution :
Anti-derivative of $\cos 3x$ is a function of $x$ whose derivative is $\cos 3x$.
It is known that,
$$\frac{d}{dx}(\sin 3x) = 3 \cos 3x$$
Therefore,
$$\cos 3x = \frac{1}{3} \frac{d}{dx}(\sin 3x)$$
or
$$\cos 3x = \frac{d}{dx}\left(\frac{1}{3} \sin 3x\right)$$
Hence, the anti-derivative of $\cos 3x$ is
$$\frac{1}{3} \sin 3x + C$$
NCERT Question.3 : Finding the Anti-Derivative (Integral) by Inspection of function $e^{2x}$.
Solution :
Anti-derivative of $e^{2x}$ is a function of $x$ whose derivative is $e^{2x}$.
It is known that,
$$\frac{d}{dx}(e^{2x}) = 2 e^{2x}$$
Therefore,
$$e^{2x} = \frac{1}{2} \frac{d}{dx}(e^{2x})$$
or
$$e^{2x} = \frac{d}{dx}\left(\frac{1}{2} e^{2x}\right)$$
Hence, the anti-derivative of $e^{2x}$ is
$$\frac{1}{2} e^{2x} + C$$
NCERT Question.4 : Finding the Anti-Derivative (Integral) by Inspection of function $(ax + b)^2$.
Solution :
Anti-derivative of $(ax + b)^2$ is a function of $x$ whose derivative is $(ax + b)^2$.
It is known that,
$$\frac{d}{dx}((ax + b)^3) = 3a (ax + b)^2$$
Therefore,
$$(ax + b)^2 = \frac{1}{3a} \frac{d}{dx}((ax + b)^3)$$
or
$$(ax + b)^2 = \frac{d}{dx}\left(\frac{1}{3a}(ax + b)^3\right)$$
Hence, the anti-derivative of $(ax + b)^2$ is
$$\frac{1}{3a}(ax + b)^3 + C$$
NCERT Question.5 : Finding the Anti-Derivative (Integral) by Inspection of function $\sin 2x – 4 e^{3x}$.
Solution :
Anti-derivative of $\sin 2x – 4 e^{3x}$ is a function of $x$ whose derivative is $\sin 2x – 4 e^{3x}$.
We know,
$$\frac{d}{dx}(\cos 2x) = -2 \sin 2x$$
Dividing by $-2$,
$$\frac{d}{dx}\left(-\frac{1}{2} \cos 2x\right) = \sin 2x \quad \text{…(i)}$$
Again,
$$\frac{d}{dx}(e^{3x}) = 3 e^{3x}$$
So,
$$\frac{d}{dx}\left(\frac{1}{3} e^{3x}\right) = e^{3x}$$
Multiplying by $-4$,
$$\frac{d}{dx}\left(-\frac{4}{3} e^{3x}\right) = -4 e^{3x} \quad \text{…(ii)}$$
Adding (i) and (ii),
$$\frac{d}{dx}\left(-\frac{1}{2} \cos 2x – \frac{4}{3} e^{3x}\right) = \sin 2x – 4 e^{3x}$$
Hence, the anti-derivative of $\sin 2x – 4 e^{3x}$ is
$$-\frac{1}{2} \cos 2x – \frac{4}{3} e^{3x} + C$$
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NCERT Question 6: Find the Integral
$$\int (4e^{3x} + 1)dx$$
Solution:
$$
\int (4e^{3x} + 1)dx
= 4\int e^{3x}dx + \int 1dx
$$
Now,
$$
\int e^{3x}dx = \frac{e^{3x}}{3}
$$
Substituting this, we get:
$$
4\int e^{3x}dx + \int 1dx = 4\left(\frac{e^{3x}}{3}\right) + x + C
$$
Simplifying,
$$
\boxed{\frac{4}{3}e^{3x} + x + C}
$$
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NCERT Question 7: Find the Integral
$$\int x^2 \left(1 – \frac{1}{x^2}\right)dx$$
Solution:
$$
\int x^2 \left(1 – \frac{1}{x^2}\right)dx
= \int (x^2 – 1)dx
$$
$$
= \int x^2dx – \int 1dx
$$
$$
= \frac{x^3}{3} – x + C
$$
Hence,
$$
\boxed{\frac{x^3}{3} – x + C}
$$
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NCERT Question 8: Find the Integral
$$\int (ax^2 + bx + c)dx$$
Solution:
$$
\int (ax^2 + bx + c)dx
= a\int x^2dx + b\int xdx + c\int 1dx
$$
$$
= a\left(\frac{x^3}{3}\right) + b\left(\frac{x^2}{2}\right) + cx + C
$$
Simplifying,
$$
\boxed{\frac{a x^3}{3} + \frac{b x^2}{2} + c x + C}
$$
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NCERT Question 9: Find the Integral
$$\int (2x^2 + e^x)\;dx$$
Solution:
$$
\int (2x^2 + e^x)\;dx
= 2\int x^2\;dx + \int e^x\;dx
$$
$$
= 2\left(\frac{x^3}{3}\right) + e^x + C
$$
Simplifying,
$$
\boxed{\frac{2x^3}{3} + e^x + C}
$$
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NCERT Question 10: Find the Integral
$$\int \left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^2\;dx$$
Solution:
Expanding the square,
$$
\left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^2 = x + \frac{1}{x} – 2
$$
Therefore,
$$
\int \left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^2\;dx
= \int \left(x + \frac{1}{x} – 2\right)\;dx
$$
$$
= \int x\;dx + \int \frac{1}{x}\;dx – 2\int 1\;dx
$$
$$
= \frac{x^2}{2} + \ln|x| – 2x + C
$$
Hence,
$$
\boxed{\frac{x^2}{2} + \ln|x| – 2x + C}
$$
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NCERT Question 11: Find the Integral
$$\int \frac{x^3 + 5x^2 – 4}{x^2}dx$$
Solution:
Simplify the integrand:
$$
\frac{x^3 + 5x^2 – 4}{x^2} = x + 5 – 4x^{-2}
$$
Therefore,
$$
\int \frac{x^3 + 5x^2 – 4}{x^2}dx = \int (x + 5 – 4x^{-2})dx
$$
Integrating each term,
$$
= \int xdx + 5\int 1dx – 4\int x^{-2}dx
$$
$$
= \frac{x^2}{2} + 5x – 4\left(\frac{x^{-1}}{-1}\right) + C
$$
Simplifying,
$$
\boxed{\frac{x^2}{2} + 5x + \frac{4}{x} + C}
$$
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