Anand Classes presents well-structured NCERT Solutions for Integrals Exercise 7.7 of Chapter 7 for Class 12 Mathematics, designed to help students master advanced integration techniques through detailed, step-by-step explanations. These notes follow the latest NCERT guidelines and provide clear concepts, solved examples, and accurate answers to enhance exam preparation and conceptual understanding. Click the print button to download study material and notes.

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NCERT Question 1: Evaluate the integral
$$\int \sqrt{4 – x^{2}}\;dx$$

Solution

$$\int \sqrt{4 – x^{2}}\;dx$$

Rewrite the integrand as:
$$\sqrt{4 – x^{2}} = \sqrt{2^{2} – x^{2}}$$

Use the standard formula:
$$\int \sqrt{a^{2} – x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2} – x^{2}} + \frac{a^{2}}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$$

Here, (a = 2). Substitute into the formula:

$$\int \sqrt{4 – x^{2}}\;dx = \frac{x}{2}\sqrt{4 – x^{2}} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) + C$$

Simplify:

$$\int \sqrt{4 – x^{2}}\;dx = \frac{x}{2}\sqrt{4 – x^{2}} + 2\sin^{-1}\left(\frac{x}{2}\right) + C$$

Final Answer

$$\boxed{\frac{x}{2}\sqrt{4 – x^{2}} + 2\sin^{-1}\left(\frac{x}{2}\right) + C}$$

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NCERT Question 2: Evaluate the integral
$$\int \sqrt{1 – 4x^2}\;dx$$

Solution

We rewrite the integral as
$$\int \sqrt{1 – (2x)^2}\;dx$$

Let
$$2x = y \quad\Rightarrow\quad 2\;dx = dy \quad\Rightarrow\quad dx = \frac{dy}{2}$$

So the integral becomes
$$\frac{1}{2}\int \sqrt{1 – y^2}\;dy$$

Using the standard result
$$\int \sqrt{1 – y^2}\;dy = \frac{y}{2}\sqrt{1 – y^2} + \frac{1}{2}\sin^{-1}(y) + C$$

we get
$$\frac{1}{2}\int \sqrt{1 – y^2}\;dy=\frac{1}{2}\left[\frac{y}{2}\sqrt{1 – y^2} + \frac{1}{2}\sin^{-1}(y)\right] + C$$

Now substitute $(y = 2x)$:
$$\int \sqrt{1 – (2x)^2}\;dx=\frac{1}{2}\left[\frac{2x}{2}\sqrt{1 – (2x)^2} + \frac{1}{2}\sin^{-1}(2x)\right] + C$$

Simplifying,
$$\int \sqrt{1 – (2x)^2}\;dx=\frac{x}{2}\sqrt{1 – 4x^2} + \frac{1}{4}\sin^{-1}(2x) + C$$

Final Answer
$$\boxed{\frac{x}{2}\sqrt{1 – 4x^2} + \frac{1}{4}\sin^{-1}(2x) + C}$$

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NCERT Question.3: Evaluate the integral
$$\int \sqrt{x^2 + 4x + 6}\;dx$$

Solution

$$\int \sqrt{x^2 + 4x + 6}\;dx$$

Rewrite the expression inside the root:

$$x^2 + 4x + 6 = (x+2)^2 + (\sqrt{2})^2$$

So the integral becomes
$$\int \sqrt{(x+2)^2 + (\sqrt{2})^2}\;dx$$

Let
$$(x+2)=y \quad\Rightarrow\quad dx = dy$$

Thus,
$$\int \sqrt{y^2 + (\sqrt{2})^2}\;dy$$

Using the standard formula
$$\int \sqrt{y^2 + a^2}\;dy = \frac{y}{2}\sqrt{y^2 + a^2} + \frac{a^2}{2}\ln\left|y + \sqrt{y^2 + a^2}\right| + C,$$

we get
$$\int \sqrt{y^2 + (\sqrt{2})^2}\;dy = \frac{y}{2}\sqrt{y^2 + 2} + \ln\left|y + \sqrt{y^2 + 2}\right| + C.$$

Substitute back $(y = x+2)$:

$$\int \sqrt{x^2 + 4x + 6}\;dx=\frac{x+2}{2}\sqrt{(x+2)^2 + 2} + \\[10pt]+\;\ln\left|x+2 + \sqrt{(x+2)^2 + 2}\right| + C.$$

Since
$$(x+2)^2 + 2 = x^2 + 4x + 6,$$

Final Answer
$$\boxed{\frac{x+2}{2}\sqrt{x^2 + 4x + 6} + \ln\left|x+2 + \sqrt{x^2 + 4x + 6}\right| + C}$$

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NCERT Question 4: Evaluate the integral
$$\int \sqrt{x^{2}+4x+1}\;dx$$

Solution

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

Complete the square inside the root:
$$x^{2}+4x+1=(x+2)^{2}-3.$$

Put
$$y=x+2\quad\Rightarrow\quad dy=dx.$$

The integral becomes
$$\int \sqrt{y^{2}-3}\;dy.$$

Use the standard formula
$$\int \sqrt{y^{2}-a^{2}}\;dy=\frac{y}{2}\sqrt{y^{2}-a^{2}}-\frac{a^{2}}{2}\ln!\bigl|y+\sqrt{y^{2}-a^{2}}\bigr|+C.$$
Here $a^{2}=3$, so
$$\int \sqrt{y^{2}-3}\;dy=\frac{y}{2}\sqrt{y^{2}-3}- \frac{3}{2}\ln!\bigl|y+\sqrt{y^{2}-3}\bigr|+C.$$

Substitute back $y=x+2$:
$$\int \sqrt{x^{2}+4x+1}\;dx
=\frac{x+2}{2}\sqrt{(x+2)^{2}-3}\;-\\[5mm]-\frac{3}{2}\ln!\bigl|x+2+\sqrt{(x+2)^{2}-3}\bigr|+C.$$

Since $(x+2)^{2}-3=x^{2}+4x+1$, the result is

Final Answer

$$\boxed{\;\displaystyle \frac{x+2}{2}\sqrt{x^{2}+4x+1}\;-\;\frac{3}{2}\ln!\bigl|x+2+\sqrt{x^{2}+4x+1}\bigr|+C\;}$$

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NCERT Question 5: Evaluate the integral
$$\displaystyle \int \sqrt{1-4x-x^{2}}\;dx$$

Solution

$$\displaystyle \int \sqrt{1-4x-x^{2}}\;dx$$

Complete the square inside the root.
$$1-4x-x^{2}=-(x^{2}+4x-1)= -\bigl((x+2)^{2}-5\bigr)=5-(x+2)^{2}$$

So the integral becomes
$$\displaystyle \int \sqrt{5-(x+2)^{2}}\;dx$$

Put
$$u=x+2\quad\Rightarrow\quad du=dx$$

Then the integral is
$$\displaystyle \int \sqrt{5-u^{2}}\;du$$

Use the standard formula
$$\displaystyle \int \sqrt{a^{2}-u^{2}}\;du=\frac{u}{2}\sqrt{a^{2}-u^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{u}{a}+C$$
with $a=\sqrt{5}$. Thus

$$\displaystyle \int \sqrt{5-u^{2}}\;du
=\frac{u}{2}\sqrt{5-u^{2}}+\frac{5}{2}\sin^{-1}\frac{u}{\sqrt{5}}+C$$

Substitute back $u=x+2$:

$$\displaystyle \int \sqrt{1-4x-x^{2}}\;dx
=\frac{x+2}{2}\sqrt{5-(x+2)^{2}}+\frac{5}{2}\sin^{-1}\frac{x+2}{\sqrt{5}}+C$$

$$\displaystyle \int \sqrt{1-4x-x^{2}}\;dx
=\frac{x+2}{2}\sqrt{1-4x-x^{2}}+\frac{5}{2}\sin^{-1}\frac{x+2}{\sqrt{5}}+C$$

Final Answer

$$\boxed{\displaystyle \frac{x+2}{2}\sqrt{1-4x-x^{2}}+\frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C}$$

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NCERT Question 6: Evaluate the integral
$$\displaystyle \int \sqrt{x^2 + 4x – 5}\; dx$$

Solution

$$\displaystyle \int \sqrt{x^2 + 4x – 5}\; dx$$

Step 1: Rewrite the expression

Complete the square:

$$x^2 + 4x – 5 = x^2 + 4x + 4 – 4 – 5$$

$$= (x+2)^2 – 9$$

So the integral becomes:

$$\int \sqrt{(x+2)^2 – 9}\; dx$$

Step 2: Substitute

Let
$$y = x + 2 \quad \Rightarrow \quad dx = dy$$

Then:

$$\int \sqrt{y^2 – 9}\; dy$$

Step 3: Use the standard formula

For
$$\int \sqrt{y^2 – a^2}\; dy=\frac{y}{2}\sqrt{y^2 – a^2} \;-\; \frac{a^2}{2}\ln\left|y + \sqrt{y^2 – a^2}\right| + C$$

Here $a^2 = 9$ so $a = 3$.

Applying:

$$\int \sqrt{y^2 – 9}\; dy = \frac{y}{2}\sqrt{y^2 – 9} \;-\; \frac{9}{2}\ln\left|y + \sqrt{y^2 – 9}\right| + C$$

Step 4: Substitute back $y = x + 2$

$$\displaystyle \int \sqrt{x^2 + 4x – 5}\; dx = \frac{x+2}{2}\sqrt{(x+2)^2 – 9} \;-\; \frac{9}{2}\ln\left|x+2 + \sqrt{(x+2)^2 – 9}\right| + C$$

Since
$$(x+2)^2 – 9 = x^2 + 4x – 5,$$

$$\displaystyle \int \sqrt{x^2 + 4x – 5}\; dx = \frac{x+2}{2}\sqrt{x^2 + 4x – 5} \;-\; \frac{9}{2}\ln\left|x+2 + \sqrt{x^2 + 4x – 5}\right| + C$$

the final answer becomes:

$$\boxed{\frac{x+2}{2}\sqrt{x^2 + 4x – 5} \;-\; \frac{9}{2}\ln\left|x + 2 + \sqrt{x^2 + 4x – 5}\right| + C}$$

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NCERT Question 7: Evaluate the integral
$$\int \sqrt{1+3x-x^{2}}\;dx$$

Solution

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

Complete the square inside the radicand:
$$1+3x-x^{2}= -\bigl(x^{2}-3x-1\bigr)
= -\Bigl((x-\frac{3}{2})^{2}-\frac{13}{4}\Bigr)
=\frac{13}{4}-(x-\frac{3}{2})^{2}.$$

Put
$$y=x-\frac{3}{2}$$
so that
$$dy=dx$$
and the integral becomes
$$\int \sqrt{\frac{13}{4}-y^{2}}\;dy.$$

Use the standard formula
$$\int \sqrt{a^{2}-y^{2}}\;dy=\frac{y}{2}\sqrt{a^{2}-y^{2}}+\frac{a^{2}}{2}\sin^{-1}\left(\frac{y}{a}\right)+C.$$

Here $a^{2}=\dfrac{13}{4}$ so $a=\dfrac{\sqrt{13}}{2}$.

Applying the formula:
$$\int \sqrt{\frac{13}{4}-y^{2}}\;dy
=\frac{y}{2}\sqrt{\frac{13}{4}-y^{2}}+\frac{13}{8}\sin^{-1}\left(\frac{2y}{\sqrt{13}}\right)+C.$$

Substitute back $y=x-\dfrac{3}{2}$ and simplify the first factor:
$$\frac{y}{2}\sqrt{\dfrac{13}{4}-y^{2}}
=\frac{x-\dfrac{3}{2}}{2}\sqrt{1+3x-x^{2}}
=\frac{2x-3}{4}\sqrt{1+3x-x^{2}}.$$

Therefore the integral is

$$\boxed{\displaystyle \int \sqrt{1+3x-x^{2}}\;dx
=\frac{2x-3}{4}\sqrt{1+3x-x^{2}}+\frac{13}{8}\sin^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C}$$

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NCERT Question 8: Evaluate the integral
$$\int \sqrt{x^{2}+3x}\;dx$$

Solution

$$\int \sqrt{x^{2}+3x}\;dx$$

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

Put
$$y=x+\dfrac{3}{2}$$
so that
$$dy=dx.$$

The integral becomes
$$\int \sqrt{y^{2}-\left(\frac{3}{2}\right)^{2}}\;dy.$$

Use the standard formula
$$\int \sqrt{y^{2}-a^{2}}\;dy=\frac{y}{2}\sqrt{y^{2}-a^{2}}-\frac{a^{2}}{2}\ln\left|y+\sqrt{y^{2}-a^{2}}\right|+C.$$

Here
$$a=\dfrac{3}{2},\quad a^{2}=\dfrac{9}{4}.$$

Applying the formula:

$$\int \sqrt{y^{2}-\left(\frac{3}{2}\right)^{2}}\;dy
=\frac{y}{2}\sqrt{y^{2}-\frac{9}{4}}-\frac{9}{8}\ln\left|y+\sqrt{y^{2}-\frac{9}{4}}\right|+C.$$

Now substitute back $y=x+\dfrac{3}{2}$:

$$\frac{y}{2}\sqrt{y^{2}-\frac{9}{4}}
=\frac{x+\dfrac{3}{2}}{2}\sqrt{x^{2}+3x}
=\frac{2x+3}{4}\sqrt{x^{2}+3x}.$$

Thus the integral is:

$$\boxed{\displaystyle
\int \sqrt{x^{2}+3x}\;dx
=\frac{2x+3}{4}\sqrt{x^{2}+3x}-\frac{9}{8}\ln\left|x+\frac{3}{2}+\sqrt{x^{2}+3x}\right|+C}$$

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NCERT Question 9: Evaluate the integral
$$\int \sqrt{1+\frac{x^{2}}{9}}\;dx$$

Solution

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

Rewrite the expression:
$$\sqrt{1+\frac{x^{2}}{9}}
=\sqrt{\frac{x^{2}+9}{9}}
=\frac{1}{3}\sqrt{x^{2}+9}.$$

Thus,
$$\int \sqrt{1+\frac{x^{2}}{9}}\;dx
=\frac{1}{3}\int \sqrt{x^{2}+9}\;dx.$$

Use the standard formula
$$\int \sqrt{x^{2}+a^{2}}\;dx
=\frac{x}{2}\sqrt{x^{2}+a^{2}}
+\frac{a^{2}}{2}\ln\left|x+\sqrt{x^{2}+a^{2}}\right|+C.$$

Here
$$a^{2}=9,\quad a=3.$$

Apply the formula:

$$\frac{1}{3}\int \sqrt{x^{2}+9}\;dx=\frac{1}{3}\left[\frac{x}{2}\sqrt{x^{2}+9}
+\frac{9}{2}\ln\left|x+\sqrt{x^{2}+9}\right|\right]+C.$$

Simplify:
$$\frac{1}{3}\int \sqrt{x^{2}+9}\;dx=\frac{x}{6}\sqrt{x^{2}+9}
+\frac{3}{2}\ln\left|x+\sqrt{x^{2}+9}\right|+C.$$

Therefore,

$$\boxed{
\int \sqrt{1+\frac{x^{2}}{9}}\;dx
=\frac{x}{6}\sqrt{x^{2}+9}
+\frac{3}{2}\ln\left|x+\sqrt{x^{2}+9}\right|+C}$$

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NCERT Question 10: Evaluate the integral
$$\int \sqrt{1+x^{2}}\;dx$$
(A) $\frac{x}{2}\sqrt{1+x^{2}}+\frac{1}{2}\ln|x+\sqrt{1+x^{2}}|+C$ $\\$
(B) $\displaystyle \frac{2}{3}(1+x^{2})^{3/2}+C$ $\\$
(C) $\displaystyle \frac{2}{3}x(1+x^{2})^{3/2}+C$ $\\$
(D) $\displaystyle \frac{x^{2}}{2}\sqrt{1+x^{2}}+\frac{1}{2}x^{2}\ln|x+\sqrt{1+x^{2}}|+C$

Solution

$$\int \sqrt{1+x^{2}}\;dx$$

Using the standard formula
$$\int \sqrt{x^{2}+a^{2}}\;dx=\frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\ln\left|x+\sqrt{x^{2}+a^{2}}\right|+C,$$
here $(a^{2}=1).$

So,
$$\int \sqrt{1+x^{2}}\;dx
=\frac{x}{2}\sqrt{1+x^{2}}
+\frac{1}{2}\ln\left|x+\sqrt{1+x^{2}}\right|+C.$$

Thus, the correct answer is Option (A).

$$\boxed{\frac{x}{2}\sqrt{1+x^{2}}+\frac{1}{2}\ln|x+\sqrt{1+x^{2}}|+C}$$

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NCERT Question 11: Evaluate the integral
$$\int \sqrt{x^{2}-8x+7}\;dx$$
(A) $\displaystyle \frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}+9\ln|x-4+\sqrt{x^{2}-8x+7}|+C$ $\\$
(B) $\displaystyle \frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}+9\ln|x+4+\sqrt{x^{2}-8x+7}|+C$ $\\$
(C) $\displaystyle \frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}-3\sqrt{2}\ln|x-4+\sqrt{x^{2}-8x+7}|+C$ $\\$
(D) $\displaystyle \frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}-\frac{9}{2}\ln|x-4+\sqrt{x^{2}-8x+7}|+C$ $\\$

Solution

$$\int \sqrt{x^{2}-8x+7}\;dx$$

Completing the square:
$$x^{2}-8x+7=(x-4)^{2}-9.$$

Let
$$y=x-4 \;dx=dy.$$

Then,
$$\int \sqrt{x^{2}-8x+7}\;dx=\int \sqrt{y^{2}-3^{2}}\;dy.$$

Using the standard formula
$$\int \sqrt{y^{2}-a^{2}}\;dy=\frac{y}{2}\sqrt{y^{2}-a^{2}}-\frac{a^{2}}{2}\ln\left|y+\sqrt{y^{2}-a^{2}}\right|+C,$$

with $(a=3)$:

$$\int \sqrt{y^{2}-3^{2}}\;dy=\frac{y}{2}\sqrt{y^{2}-9}-\frac{9}{2}\ln|y+\sqrt{y^{2}-9}|+C.$$

Substituting back $(y=x-4)$:

$$\int \sqrt{x^{2}-8x+7}\;dx=\frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}-\\[0.75em] -\frac{9}{2}\ln|x-4+\sqrt{x^{2}-8x+7}|+C$$

The final Answer :

$$\boxed{\frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}-\frac{9}{2}\ln|x-4+\sqrt{x^{2}-8x+7}|+C}$$

Thus, the correct answer is Option (D).

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